Problem
(Exercise 15.12 in Bott & Tu's Differential Forms in Algebraic Topology) If $X$ is a space having a good cover, e.g., a triangularizable space, and $Y$ is any topological space, prove using the spectral sequence of the fiber bundle $\pi\colon X\times Y\to X$ that $H^n(X\times Y)=\bigoplus_{p+q=n}H^p(X,H^q(Y))$.
Remark
Cochain complexes and cohomology groups are with integer coefficients.
Thoughts
Suppose $\mathfrak U$ is a good cover of $X$, and $\pi^{-1}\mathfrak U$ is a cover of $X\times Y$. Consider the double complex $K^{\bullet,\bullet}=C^\bullet(\pi^{-1}\mathfrak U,S^\bullet(-\times Y))$, where $S^\bullet$ is the singular cochain complex. On one hand, consider the spectral sequence $E_r'$ associated to filtration by rows of $K^{\bullet,\bullet}$ which is $E_2'$ degenerate, we have $$G'H_D^n(K)=\bigoplus_{p+q=n}E_2^{\prime p,q}=H_S^n(X\times Y)$$ to be the singular cohomology group.
Note that the filtration of $H_D^n$ should be $\underbrace{H_D^n\supseteq0}_{E_2^{0,n}}$, therefore $H_D^n(K)=H_S^n(X\times Y)$.
On the other hand, consider the spectral sequence $E_r$ associated to filtration by columns of $K^{\bullet,\bullet}$, we have $E_2^{p,q}=H^p(\pi^{-1}\mathfrak U,\mathscr H^q)=H^p(X,H^q(Y))$ where $\mathscr H^q$ is the presheaf $U\mapsto H^q(\pi^{-1}U)$.
However, I cannot see any reason either that $E_r$ is $E_2$ degenerate or that $GH_D^n(K)=H_D^n(K)$ for filtration of columns. I need to know what to do next.
Any idea? Thanks!
Update
It seems that $E_2$-degeneracy is easier, although for the moment I don't have a rigorous argument for this.
Roughly speaking, an element $e^{pq}\in E_2^{pq}\cong H^p(\mathfrak U,H^q(Y))$, represented by some $c^{pq}\in C^p(\mathfrak U,S^q(-\times Y))$. We note that for every "good" $U$, the morphism $\phi_U\colon S^\bullet(Y)\to S^\bullet(U\times Y)$ induced by the projection $U\times Y\to Y$ is a quasi-isomorphism. On the other hand, the morphisms $\phi_U$ is natural in $U$, i.e. they constitute a natural transformation between the constant functor $U\mapsto S^q(Y)$ and the contravariant functor $U\mapsto S^\bullet(U\times Y)$. We can identify $c^{pq}$ with an element in $C^p(\mathfrak U,S^q(Y))$ with some effort. We can then extend $c^{pq}$ throgh the zig-zag path indefinitely. For example, the image of $\delta c^{pq}\in C^{p+1}(\mathfrak U,S^q(Y))$ in $C^{p+1}(\mathfrak U,H^q(Y))$ is zero, hence there exists a $c_2^{p+1,q-1}\in C^{p+1}(\mathfrak U,C^{q-1}(Y))$ such that $dc_2^{p+1,q-1}=c^{p+1,q}$, and we can further extend $c_2^{p+1,q-1}\in C^{p+1}(\mathfrak U,C^{q-1}(Y))$ to some $c_3^{p+2,q-2}\in C^{p+2}(\mathfrak U,C^{q-2}(Y))$, etc.
I still have no idea about the extension problem.