 Solution

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis 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. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.