= Solution
Use the standard <CW complex> structure on the <real projective plane>, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the <cellular chain complex> has $de_2=2e_1$ and $de_1=0$.
The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The <cellular chains of a product of finite CW complexes> have boundary $d(e_i\times e_j)=de_i\times e_j+(-1)^ie_i\times de_j$. In the ordered bases
$$
C_1=(e_{10},e_{01}),\quad C_2=(e_{20},e_{11},e_{02}),\quad C_3=(e_{21},e_{12}),\quad C_4=(e_{22}),
$$
the nonzero matrices are
$$
d_2=\begin{pmatrix}2&0&0\\0&0&2\end{pmatrix},\qquad
d_3=\begin{pmatrix}0&0\\2&-2\\0&0\end{pmatrix},\qquad
d_4=\begin{pmatrix}2\\2\end{pmatrix},\qquad d_1=0.
$$
In particular $d_2d_3=d_3d_4=0$. The degree-one <homology> is $\mathbb Z^2/2\mathbb Z^2$. The degree-two <cellular cycles> are $\mathbb Ze_{11}$, and the <cellular boundaries> there are $2\mathbb Ze_{11}$. The degree-three cycles are $\mathbb Z(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the <integral homology of two real projective planes>:
$$
\boxed{H_r(\mathbb{RP}^2\times\mathbb{RP}^2;\mathbb Z)=
\begin{cases}\mathbb Z,&r=0,\\(\mathbb Z/2)^2,&r=1,\\\mathbb Z/2,&r=2,3,\\0,&\text{otherwise}.\end{cases}}
$$
The degree-three torsion is the <Tor functor> contribution in the <Künneth theorem>; it would be missed by retaining only tensor products of the two factors' homology groups.
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