= Solution
If $X$ is <Hausdorff>, every pair $x\ne y$ has disjoint open neighborhoods $U$ and $V$. The rectangle $U\times V$ is an open neighborhood of $(x,y)$ disjoint from the diagonal. Hence every point of its complement is interior, so the diagonal is a <closed set>.
Conversely, if the diagonal is closed, its complement is open in the <product topology>. For $x\ne y$, the rectangle-basis property gives open $U\ni x$ and $V\ni y$ with $U\times V$ disjoint from the diagonal. If $z\in U\cap V$, then $(z,z)$ lies in that rectangle, a contradiction. Thus $U\cap V=\varnothing$, giving the <Hausdorff> separation condition. Consequently
$$
\boxed{X\text{ is Hausdorff}\iff\Delta_X\text{ is closed in }X\times X.}
$$
The argument includes spaces with zero or one point, for which both conditions hold.
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