On the $H^{\ast}$-module structure on homology group To see $H_{\ast}(X)$ is actually an $H^\ast$-module, one of the fact we need to prove is
$$a\cap(b\cap z)=(a\cup b)\cap z$$
for $a,b\in H^*(X)$ and $z\in H_*X$.
Here we adopt the definition of cap product in Davis-Kirk p.64, which use the Eilenberg-Zilber map 
$$A: S_*(X\times X)\rightarrow S_*X\otimes S_*X $$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.
 A: 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
$$H_n(X;A)\cong\pi_n^S(X\wedge HA)=\{S^n,X\wedge HA\},\qquad H^n(X;A)\cong \{X,\Sigma^nHA\}.$$
Here $HA$ is the Eilenberg-Mac Lane ring spectrum associated to the abelian group $A$, with multiplication $\mu:HA\wedge HA\rightarrow HA$. The curly brackets $\{S^n,X\wedge HA\}$ 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 $\alpha\in H^m(X;A)$ and $\beta\in H^n(X;A)$ is the class $\alpha\cup\beta\in H^{n+m}(X;A)$ represented by the composition
$$\alpha\cup\beta:X\xrightarrow{\Delta} X\wedge X\xrightarrow{\alpha\cup\beta} \Sigma^mHA\wedge \Sigma^mHA\xrightarrow{\cong}\Sigma^{m+n}HA\wedge HA\xrightarrow{\Sigma^{m+n}\mu} \Sigma^{m+n}HA.$$
On the other hand, given $\alpha\in H^m(X;A)$ and $x\in H_k(X;A)$, their cap product is represented by the composition
$$\alpha\cap x:S^{k-m}\cong S^{-m}\wedge S^k\xrightarrow{1\wedge x}  S^{-m}\wedge X\wedge HA\xrightarrow{1\wedge\Delta\wedge 1} S^{-m}\wedge X\wedge X\wedge HA\xrightarrow{1\wedge 1\wedge\alpha\wedge 1} S^{-m}\wedge X\wedge \Sigma^mHA\wedge HA\cong X\wedge HA\wedge HA\xrightarrow{1\wedge\mu}  X\wedge HA,$$
and is a class in $\{S^{k-m},X\wedge HA\}\cong H_{n-k}(X;A)$. Note that the negative dimensional spheres make sense in the category of spectra.
Now if $\beta\in H^n(X;A)$ is another cohomology class, consider the diagram
$\require{AMScd}$
\begin{CD}
    S^{k-m-n}@>1\wedge x>>S^{-(n+m)}\wedge X\wedge HA @>1\wedge\Delta\wedge 1>>S^{-(n+m)}\wedge X\wedge X\wedge HA@>1\wedge 1\wedge\alpha\cup \beta\wedge 1>>X\wedge HA\wedge HA\\
    @VV 1\wedge 1\wedge xV   @VV 1\wedge \Delta\wedge 1 V @V V 1\wedge 1\wedge \Delta\wedge 1V @VV1\wedge\mu V\\
    S^{-n}\wedge S^{-m}\wedge X\wedge HA @>1\wedge1\wedge\Delta\wedge 1>> S^{-(n+m)}\wedge X\wedge X\wedge HA@>1\wedge\Delta\wedge 1\wedge 1>>S^{-(n+m)}\wedge X\wedge X\wedge X\wedge HA@>1\wedge\mu(\alpha\wedge\beta)\wedge 1>> X\wedge HA\\
@VV 1\wedge 1\wedge \Delta \wedge 1V@VV1\wedge \Delta\wedge\beta\wedge 1V@VV1\wedge\alpha\wedge \beta\wedge 1V@VV=V\\
S^{-n}\wedge S^{-m}\wedge X\wedge X\wedge HA@>1\wedge1\wedge\Delta\wedge \beta\wedge1>>S^{-n}\wedge X\wedge X\wedge HA\wedge HA@>1\wedge1\wedge\alpha\wedge 1>>X\wedge HA\wedge HA\wedge HA@>1\wedge\mu\wedge1 >>X\wedge HA
\end{CD}
When we use the equalities $\mu(1\wedge \mu)=\mu(\mu\wedge 1)$ and $(1\wedge \Delta)\Delta=(\Delta\wedge 1)\Delta$ (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 $(\alpha\cup\beta)\cap x$, whilst the clockwise composition around the diagram represents $\alpha\cap(\beta\cap x)$. Since the diagram commutes, we see that
$$(\alpha\cup\beta)\cap x=\alpha\cap(\beta\cap x)$$
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.
