proof for Kunneth formula using the double complex prove: When a homomorphism $f: K \rightarrow K'$ of double complexes induces $H_d$-isomorphism, it also induces $H_D$-isomorphism.
This is in the context of proving Kunneth formula using the double complex. 
The map is the pullback of wedge products:
Let $U=\{U_{\alpha}\}$ be a good cover for $M$ and $\pi: M\times F\rightarrow M$ projection into $M$ and $\rho$ likewise into $F$. Then $\{\pi^{-1}U_{\alpha}\}$ is some cover for $ M\times F$. Assume cohomology $H^*(F)$ is finite dimensional with closed forms basis $\{[\omega_{\alpha}]\}$ Then define 
$$ \pi^*: H^*(F) \otimes C^*(U,\Omega^*)\rightarrow C^*(\pi^{-1}U,\Omega^*)$$
$$ \pi^*([\omega_{\alpha}]\otimes\phi)=\rho^*\omega_{\alpha}\wedge \pi^*\phi$$
This commutes with $d$ and $\delta$ and induces an isomorphism  in $d$-cohomology. 
*without use of spectral sequences which comes later in the book. 
book bott and Tu: 108 and 107, http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf
Edit: I do not understand the answer below. The cohomolgy classes for me are given by "diagonals, see text. The answer below does not seem correct to me.
 A: Let $K$ and $K'$ be two first-quadrant cohomological complexes, let $d$ be the vertical differentials and $D$ the total one. Let $f:K\to K'$ be a morphism which induces an isomorphism in  cohomology with respect to $d$.
Let us suppose that both $K$ and $K'$ have finitely many nonzero columns, and let us do induction on the number of columns. The hypothesis holds in your case, since the de Rham complex of a manifold has finite length.
Let $K_1$ and $K'_1$ be the subcomplexes of $K$ and $K'$ of all columns of positive degree and let $C$ and $C'$ be the first columns of $K$ and of $K'$. There is a commutatvie diagram of complexes
0 —> K_1  —> K  —> C  —> 0
     |       |     |
     V       V     V
0 —> K_1' —> K' —> C' —> 0

with exact rows and vertical arrows induced by $f$, which induces an infinite ladder of the form
H_{i-1}(C)  —> H_i(K_1)  —> H_i(K)  —> H_i(C)  —> H+i(K_1)   
     |         |            |          |          |
     V         V            V          V          V
H_{i-1}(C') —> H_i(K_1') —> H_i(K') —> H_i(C') —> H+i(K_1') 

Here the vertical arrows $H_{i-1}(C)\to H_{i-1}(C')$ and $H_{i}(C)\to H_{i}(C')$ are isomorphism by hypothesis, and the arrows $H_i(K_1)\to H_i(K_1')$ and $H_i(K_1)\to H_i(K_1')$ are isomorphisms inductively, as $K_1$ and $K_1$ have one less nonzero column than $K$ and $K'$. The arrow $H_i(K)\to H_i(K')$ is an isomorphism by the $5$-lemma.

The result is true even for complexes with infinitely many nonzero arrows. I'll leave it for someone with the energy to do so to write a proof :-)
