Cfrgtkky

To see $H_{ast}(X)$ is actually an $H^ast$-module, one of the fact we need to prove is
$$acap(bcap z)=(acup b)cap z$$
for $a,bin H^*(X)$ and $zin H_*X$.

Here we adopt the definition of cap product in Davis-Kirk p.64, which use the Eilenberg-Zilber map
$$A: S_*(Xtimes X)rightarrow S_*Xotimes S_*X $$This makes the prove work difficult since we actually do not know the map $A$ explicitly. Instead, we only know it is functorial, which I doubt is useless here.

I know there is a simple proof using another construction with the Alexander-Whitney diagonal approximation. But I am still curious about if we have a proof without it.

I'm not sure you will count this as a complete answer, but here is a (stable) homotopy theoretic approach to the problem. Recall that by Brown representability, for a CW complex $X$ and an abelian group $A$ there are isomorphisms of groups

$$H_n(X;A)congpi_n^S(Xwedge HA)={S^n,Xwedge HA},qquad H^n(X;A)cong {X,Sigma^nHA}.$$

Here $HA$ is the Eilenberg-Mac Lane ring spectrum associated to the abelian group $A$, with multiplication $mu:HAwedge HArightarrow HA$. The curly brackets ${S^n,Xwedge HA}$ represent homotopy classes of stable maps in the category of spectra, and I am supressing the (infinite) suspension functor from notation (so, for instance, I should write $Sigma^infty X$ for the image of $X$ in the category of spectra, but do not for readability).

If $X$ is not a CW complex, then a suitably CW replacement $bar X$ should be taken, which should then replace $X$ in all of the previous isomorphisms.

Now the cup product of $alphain H^m(X;A)$ and $betain H^n(X;A)$ is the class $alphacupbetain H^{n+m}(X;A)$ represented by the composition

$$alphacupbeta:Xxrightarrow{Delta} Xwedge Xxrightarrow{alphacupbeta} Sigma^mHAwedge Sigma^mHAxrightarrow{cong}Sigma^{m+n}HAwedge HAxrightarrow{Sigma^{m+n}mu} Sigma^{m+n}HA.$$

On the other hand, given $alphain H^m(X;A)$ and $xin H_k(X;A)$, their cap product is represented by the composition

$$alphacap x:S^{k-m}cong S^{-m}wedge S^kxrightarrow{1wedge x} S^{-m}wedge Xwedge HAxrightarrow{1wedgeDeltawedge 1} S^{-m}wedge Xwedge Xwedge HAxrightarrow{1wedge 1wedgealphawedge 1} S^{-m}wedge Xwedge Sigma^mHAwedge HAcong Xwedge HAwedge HAxrightarrow{1wedgemu} Xwedge HA,$$

and is a class in ${S^{k-m},Xwedge HA}cong H_{n-k}(X;A)$. Note that the negative dimensional spheres make sense in the category of spectra.

Now if $betain H^n(X;A)$ is another cohomology class, consider the diagram

$require{AMScd}$
begin{CD}
S^{k-m-n}@>1wedge x>>S^{-(n+m)}wedge Xwedge HA @>1wedgeDeltawedge 1>>S^{-(n+m)}wedge Xwedge Xwedge HA@>1wedge 1wedgealphacup betawedge 1>>Xwedge HAwedge HA\
@VV 1wedge 1wedge xV @VV 1wedge Deltawedge 1 V @V V 1wedge 1wedge Deltawedge 1V @VV1wedgemu V\
S^{-n}wedge S^{-m}wedge Xwedge HA @>1wedge1wedgeDeltawedge 1>> S^{-(n+m)}wedge Xwedge Xwedge HA@>1wedgeDeltawedge 1wedge 1>>S^{-(n+m)}wedge Xwedge Xwedge Xwedge HA@>1wedgemu(alphawedgebeta)wedge 1>> Xwedge HA\
@VV 1wedge 1wedge Delta wedge 1V@VV1wedge Deltawedgebetawedge 1V@VV1wedgealphawedge betawedge 1V@VV=V\
S^{-n}wedge S^{-m}wedge Xwedge Xwedge HA@>1wedge1wedgeDeltawedge betawedge1>>S^{-n}wedge Xwedge Xwedge HAwedge HA@>1wedge1wedgealphawedge 1>>Xwedge HAwedge HAwedge HA@>1wedgemuwedge1 >>Xwedge HA
end{CD}

When we use the equalities $mu(1wedge mu)=mu(muwedge 1)$ and $(1wedge Delta)Delta=(Deltawedge 1)Delta$ (these are equalities of stables homotopy classes) we see that the diagram commutes in the homotopy category of spectra. Note that the anti-clockwise composition around the diagram from top-left to bottom-right represents $(alphacupbeta)cap x$, whilst the clockwise composition around the diagram represents $alphacap(betacap x)$. Since the diagram commutes, we see that

$$(alphacupbeta)cap x=alphacap(betacap x)$$

as required. Moreover, not one Eilenberg-Zilber nor Alexander-Whitney map was ever even thought about.

Edit: The big diagram might be a bit obscured, but I'm not sure how else I could format the thing to make it fit.