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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.

The product topology on the Cartesian product $X\times Y$ has a basis of a topology consisting of all rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the smallest topology making the two coordinate projections continuous.

Take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, use the Hausdorff condition in $X$ to find disjoint open neighbourhoods $U_1,U_2$. Then $U_1\times Y$ and $U_2\times Y$ are disjoint open neighbourhoods of the two points. If $x_1=x_2$, then $y_1\ne y_2$; disjoint open neighbourhoods $V_1,V_2$ in $Y$ give $X\times V_1$ and $X\times V_2$. Thus $X\times Y$ is Hausdorff, as established by the forward implication of the Hausdorff property of a nonempty product. An empty Cartesian product is automatically Hausdorff.