= Solution
If both factors are <Hausdorff>, two distinct product points differ in at least one coordinate. Disjoint <open neighbourhoods> in that coordinate, multiplied by the other whole factor, give disjoint <open neighbourhoods> of the product points. Thus the product is <Hausdorff>.
Conversely, assume the product is <Hausdorff>. Fix $y_0\in Y$, using nonemptiness. For $x\ne x'$, choose disjoint <open neighbourhoods> of $(x,y_0)$ and $(x',y_0)$ and refine them to basis rectangles $U\times V$ and $U'\times V'$. Since $y_0\in V\cap V'$, their disjointness forces $U\cap U'=\varnothing$. These separate $x,x'$, proving that $X$ is <Hausdorff>. Fixing a point of $X$ proves the same for $Y$.
$$
\boxed{X\times Y\text{ Hausdorff}\iff X,Y\text{ Hausdorff},\quad X,Y\ne\varnothing.}
$$
The nonempty hypothesis is essential: a product with the <empty set> is empty regardless of the other factor. This is the <Hausdorff property of a nonempty product>.
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