= Solution
Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The family
$$
\mathcal D=\{B_i\times C_j:(i,j)\in I\times J\}
$$
is countable, since the Cartesian product of two countable sets is countable. It is a <basis of a topology> for the <product topology>: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. Therefore
$$
\boxed{\text{The product has a countable basis.}}
$$
In standard terminology, a finite product of <second-countable spaces> is second-countable.
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