Cfrgtkky

There is an exercise on page 14 in the Tomas Jech's "Set Theory":

Every nonempty $X subseteq mathbb{N}$ has an $in$-minimal element.
[Pick $n in X$ and look at $X cap n$.]

and there is a good solution written by the user egreg: https://math.stackexchange.com/a/1526646/251394 . (We deal with the intersection of all elements of the set.)

This proof has a lot of inductions. For example, we need to prove trichotomy. And induction itself is not yet proved.
That is strange because other theorems nearby (1.1-1.6 and 1.8-1.9) are very easy, relatively to 1.7.

Could you please suggest a shorter proof?

p.s. We also can use the regularity axiom, but it is not fair.

p.p.s. I think, it is sufficient to show that $bigcap X in X$. (Then proof by contradiction: assume $sin X cap bigcap X$, so $sin X$ and $forall min X.sin m$, which means that $sin s$, so there is a contradiction by ex.1.5)

By exercise 1.6 you have that for every element $n$ of $mathbb{N}$, if $zsubseteq n$ is not empty, then has a $in$-minimal element. So, you have two cases

$Xcap n=emptyset$: In this case $n$ is a $in$-minimal element of $X$

$Xcap nneqemptyset$: As $Xcap nsubseteq n$ then exists a $in$-minimal element $min Xcap n$. So, $m$ is $in$-minimal in $X$ too.
(Edited: in the last sentence of the second case we use ex.1.4, which states that every natural number is transitive.)