Let $X$ be a triangulation of $[0,1]^4$ and $A$ is a simplicial subcomplex of $X$. I would like to show that the cup product $$H^2(X,A)\times H^2(X,A)\to H^4(X,A)$$ is trivial.
It is realitvely easy to show that $H^4(A)$ is trivial, which was answered here.
All examples of nontrivial cup products I could find are somehow derived from attaching 4-cells to $X^{(2)}/A^{(2)}$ in a nontrivial way, such as in the case of $\Bbb CP^2$, or $S^2\times S^2$---but it seems to me that it is impossible to realize this all within $\Bbb R^4$. Any hint?
Since $X$ is contractible, $X/A$ is naturally homotopy equivalent to the suspension $\Sigma A$. (Explicitly, contractibility of $X$ allows you to extend the inclusion $A\to X$ to a cone $CA$ on $A$, and then you get an induced map $\Sigma A=CA/A\to X/A$. From the long exact sequences on cohomology it is immediate that this map induces isomorphisms on cohomology, which is all we need here; it takes a bit more work to show it is a homotopy equivalence.) But all cup products of positive-degree cohomology classes on a suspension vanish (see here, for instance). Since $H^*(X,A)$ can be identified with the reduced cohomology of $X/A$, this answers your question.