 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 $x\ne x^{\prime }$, choose disjoint open neighbourhoods of $(x,y_0)$ and $(x^{\prime },y_0)$ and refine them to basis rectangles $U\times V$ and $U^{\prime }\times V^{\prime }$. Since $y_0\in V\cap V^{\prime }$, their disjointness forces $U\cap U^{\prime }=\varnothing$. These separate $x,x^{\prime }$, proving that $X$ is Hausdorff. Fixing a point of $X$ proves the same for $Y$.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.