Python 2, 56 bytes

for n in range(9999):

 if eval('<'.join(`n`))**n:print n

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Converts each number like 124 to an expression 1<2<4 and evaluates it to check if the digits are sorted,

A hiccup happens for one-digit numbers giving an expression that just is the number itself. This causes 0 to evaluate to a Falsey value even though it should be printed. This is fixed by a trick suggested by Erik the Outgolfer of doing **n, which gives truthy value 0**0 for n=0 and doesn't affect the truth value otherwise.

Haskell, 50 bytes

unlines[s|s<-show<$>[0..6^5],s==scanl1(max.succ)s]

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Outputs a multiline string. We check that the number s increasing using s==scanl1(max.succ)s, a variant of the usual sortedness check s==scanl1 max s that ensures strict sortedness by incrementing each digit character before taking the maximum of it and the next digit.

Ourous saved a byte by using 6^5 as the upper bound in place of a 4-digit number.

Python 2, 55 bytes

i=0

exec"print ini+=1nif eval('<'.join(`i`)):1;"*7000

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Japt -R, 12 11 8 bytes

L²Ç¶ìüÃð

Test it

L            :100

 ²           :Squared

  Ç          :Map the range [0,L²)

    ì        :  Split to a digit array

     ü       :  For the sake if simplicity*, let's say: Sort & deduplicate

             :  Implicitly rejoin to an integer

   ¶         :  Test for equality with original number

      Ã      :End map

       ð     :Get 0-based indices of truthy elements

             :Implicitly join with newlines and output

*Or, to offer a better explanation: the ü method sorts an array and splits it into equal elements (e.g., [8,4,8,4].ü() -> [[4,4],[8,8]]) and then, in what seems to be a strange quirk and hopefully not a bug, the ì method, when converting the array back to a number, takes the first element of each nested array, rather than first flattening the array, which is what I expected when I tried this trick (e.g., [[4,4],[8,8]].ì() -> 48).

Jelly, 7 bytes

9ŒPḌḣ⁹Y

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How it works

9ŒPḌḣ⁹Y  Main link. No arguments.



9        Set the return value to 9.

 ŒP      Powerset; promote 9 to [1, ..., 9] and generate all subsets.

   Ḍ     Undecimal; map the subsets of digits to the integers they represent.

     ⁹   Yield 256.

    ḣ    Dyadic head; take the first 256 elements of the integer list.

      Y  Separate the result by linefeeds.

R, 62 49 bytes

`[`=write;0[1];for(i in 1:4)combn(1:9,i)[1,i,,""]

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Because combn iterates through its input in the order given, it's easy to create all the lexicographically increasing integers, printing them out in order. write prints them each i-digit number in lines of width i, neatly fulfilling the newline requirement as well.

Haskell, 56 55 bytes

Edit: -1 byte thanks to @Ourous

mapM print$filter(and.(zipWith(<)<*>tail).show)[0..6^5]

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PowerShell, 42 40 bytes

0..1e4|?{-join("$_"|% t*y|sort -u)-eq$_}

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Loop from 0 to 1e4 (i.e., 10000). Pull out those objects where |?{...} the number as a string $_ is -equal to the number cast toCharArray and then sorted with the -unique flag. In other words, only numbers that are the same as their sorted and deduplicated strings. Each of those are left on the pipeline and output is implicit.

Perl 6, 26 bytes

.say if [<] .comb for ^1e4

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.comb produces a list of the digits of each number, and [<] does a less-than reduction on that list, equivalent to: digit1 < digit2 < ... < digitN.

J, 26 bytes

,.(#~(-:/:~@~.)@":"0)i.1e4

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explanation

,. (#~ (-: /:~@~.)@":"0) i.1e4

                         i.1e4  NB. list 0 to 9999

                     "0         NB. for each number in the input list

                  @":"0         NB. convert it to a string and

   (#~ (         )              NB. remove any not passing this test:

        -:                      NB. the string of digits matches

              @~.               NB. the nub of the digits (dups removed)

           /:~                  NB. sorted

,.                              NB. ravel items: take the result of all that

                                NB. and turn it into a big column

Common Lisp, 74 72 bytes

(dotimes(i 7e3)(format(apply'char<(coerce(format()"~d"i)'list))"~d~%"i))

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-2 bytes thank to @Shaggy!

Pyth, 10 bytes

jiRThc2yS9

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How it works

jiRThc2yS9

        S9  Yield [1, 2, 3, 4, 5, 6, 7, 8, 9].

       y    Take all 512 subsets.

     c2     Split the array of subsets into 2 pieces (256 subsets each).

    h       Head; take the first piece.

 iRT        Convert each subset from base 10 to integer.

j           Separate by newlines.

V, 41 bytes

7000ïÎaÛ

Îy$úúP

Ç^¨ä*©±$/d

ÎãlD

爱/d

HO0

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Hexdump:

00000000: 3730 3030 efce 61db 0ace 7924 fafa 500a  7000..a...y$..P.

00000010: c75e a8e4 2aa9 b124 2f64 0ace e36c 440a  .^..*..$/d...lD.

00000020: e788 b12f 640a 484f 30                   .../d.HO0

Perl 5, 55 bytes

$"=$,;map"@{[sort{$a-$b}/./g]}"-$_||/(.)1/||say,0..1E4

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Charcoal, 19 bytes

ΦＥＸχ⁴Ｉι¬Φι∧쬋§ι⊖μλ

Try it online! Link is to verbose version of code. Explanation:

   χ                Predefined variable 10

  Ｘ                 To the power

    ⁴               Literal 4

 Ｅ                  Map over implicit range

      ι             Current value

     Ｉ              Cast to string

Φ                   Filter over strings where

         ι          Current string

        Φ           Filtered over characters

           μ        Character index (is nonzero)

          ∧         And

                 μ  Character index

                ⊖   Decremented

              §     Indexed into

               ι    Current string

            ¬       Is not

             ‹      Less than

                  λ Current character

       ¬            Results in an empty string

                    Implicitly print matches on separate lines

05AB1E (legacy), 8 bytes

4°ÝD€êû

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Works in the new version of 05AB1E as well but is painfully slow for some reason.

How?

4°ÝD€êû – Full program.

4°Ý      – Push [0 ... 10000].

   Dې   РPush each integer in [0 ... 10000] sorted and deduplicated at the same time.

      û – And join the interection of the two lists by newlines.

Jelly, 13 9 8 bytes

Saved 5 bytes thanks to @Dennis

9œcⱮ4ẎŻY

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Explanation

Generates all lexicographically increasing numbers below 10000 by taking the digits [1...9] and finding all combinations of length ≤ 4.

9œcⱮ4ẎŻY    Main link. Arguments: none

9           Yield 9.

   Ɱ4       For each n in [1...4]:

 œc           Yield the combinations of the range [1...9] of length n.

     Ẏ      Tighten; dump each of the 4 lists generated into the main list.

      Ż     Prepend a 0 to the list.

       Y    Join on newlines.

Jelly, 11 10 9 bytes

Saved a byte thanks to @EriktheOutgolfer

ȷ4Ḷ<ƝẠ$ƇY

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Explanation

Filters through the range, keeping the numbers that are lexicographically increasing.

ȷ4Ḷ<ƝẠ$ƇY    Main link. Arguments: none

ȷ4           Yield 10^4 (10000).

  Ḷ          Generate the range [0...10000).

       Ƈ     Filter; yield only the numbers where this link return a truthy value.

      $        Run these two links and yield the result.

    Ɲ            For each pair of items (digits) in the number:

   <               Check whether the left digit is less than the right digit.

     Ạ           All; check that every comparison yielded true.

               This yields whether the digits are strictly increasing.

        Y    Join the filtered list on newlines.

JavaScript, 64 bytes

A bit of pub golf so probably far from optimal.

(f=n=>n&&f(n-1)+([...n+``].every(x=>y<(y=x),y=0)?`

`+n:``))(7e3)

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Yes, doing it without an IIFE would be a few bytes shorter but that throws an overflow error when called, which would normally be fine as we can assume infinite memory for the purposes of code golf but, to me, doesn't seem to be in the spirit of KC challenges.

Wolfram Language (Mathematica), 36 bytes

After I wrote this, it was clarified that each number must be on a new line, so +7 bytes for the Print/@.

This method takes advantage of the fact that the Subsets function 1) doesn't replicate any digits and 2) sorts the output by set size and set contents. FromDigits assembles each list of digits.

-1 byte thanks to @Mr.Xcoder

Print/@FromDigits/@Range@9~Subsets~4

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K (ngn/k) / K (oK), 32 30 26 bytes

Solution:

`0:$&&/'1_'>':'" ",'$!9999

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Explanation:

`0:$&&/'1_'>':'" ",'$!9999 / the solution

                     !9999 / range 0..9998 (could use !6890)

                    $      / string

               " ",'       / prepend " " to each (lower than "0" in ascii)

            >:'            / greater-than each-previous?

         1_'               / drop first result from each

      &/'                  / max (&) over (/)

    &                      / indices where true

   $                       / convert to string

`0:                        / print to stdout

Python 2, 63 bytes

for i in range(6790):

 if`i`=="".join(sorted(set(`i`))):print i

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Python 2, 61 bytes

for i in range(9999):

 if list(`i`)==sorted(set(`i`)):print i

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Python 2, 64 61 bytes

lambda:[x for x in range(9999)if sorted(set(`x`))==list(`x`)]

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Gets the unique characters of the integer's string representation, sorts them, and compares the result to the original number.

T-SQL, 188 bytes

WITH a AS(SELECT 0n UNION ALL SELECT n+1FROM a WHERE n<9)

SELECT ABS(CONCAT(a.n,b.n,c.n,d.n))

FROM a,a b,a c,a d

WHERE(a.n<b.n OR a.n+b.n=0)

AND(b.n<c.n OR b.n+c.n=0)

AND(c.n<d.n OR c.n+d.n=0)

Line breaks are for readability only.

I'm certain there must be more efficient ways to do this in SQL, but this was the first thing I thought of. Explanation:

1. Declare an in-memory table with values 0 to 9


2. Cross-join 4 copies of this table for all possible values from 0000 to 9999


3. Messy WHERE clause to ensure the digits are strictly increasing (or both 0)


4. Smash the digits together (CONCAT) and convert to integer (ABS)


5. The resulting rows may or may not be sorted, but the challenge doesn't appear to require that.

Stax, 8 bytes

¬ ▬A♥¶N∙

Run and debug it

Clean, 90 bytes

import StdEnv

Start=(0,[('

',n)\n<-[1..6^5]|(l=removeDup l==sort l)[c\c<-:toString n]])

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Red, 59 bytes

repeat n 9999[if(d: n - 1)= do sort unique form d[print d]]

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Jelly, 7 bytes

<ƝẠ$⁹#Y

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How?

<ƝẠ$⁹#Y - Main Link: no arguments (implicit z=0)

    ⁹   - literal 256

     #  - count up from n=z (0) finding the first 256 for which this is truthy:

   $    -   last two links as a monad:

 Ɲ      -     neighbours (implicitly gets digits of n):

<       -       less than?

  Ạ     -     all truthy? (N.B. yields 1 for an empty list)

      Y - join with newlines

C (gcc), 97 bytes

g(n,l,r){for(r=l=10;n;n/=10)r*=n%10<l,l=n%10;n=r;}f(i){for(i=0;i<1e4;i++)g(i)&&printf("%dn",i);}

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C# (Visual C# Interactive Compiler), 102 101 ... 73 bytes

-12 and -4 thanks @Dennis!

for(var i=0;i<7e3;i++)if((i+"").Aggregate((a,b)=>a<b?b:':')<':')Print(i);

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Each integer from 0 to 7k tested by first converting it into a string. Leveraging the fact that C# treats strings as character enumerables and LINQ, an aggregate is calculated for each character enumerable as follows:

• compare the accumulated value with the current character


• if the current character is greater than the accumulation, return the current character


• otherwise return : which is greater than 9
  

If the result of this is less than :, then the number has lexicographically increasing digits.