= Solution
Choose <countable> <dense> <sets> $D_X\subseteq X$ and $D_Y\subseteq Y$. Their <Cartesian product> is <countable>. Every nonempty basic <open set> of the <product topology> has the form $U\times V$ with $U,V$ nonempty <open> in their respective <topological spaces>. Density gives $d_X\in D_X\cap U$ and $d_Y\in D_Y\cap V$, hence $(d_X,d_Y)\in(D_X\times D_Y)\cap(U\times V)$.
Every nonempty <open set> in the product contains such a basic <open set>. Thus $D_X\times D_Y$ is <dense> in $X\times Y$, proving \b[<topological separability> of the product]. If one factor is empty the product is empty and the conclusion holds directly.
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