Theorem. If $X$ is a topological space in which every point has a closed Hausdorff neighbourhood, then $X$ is Hausdorff.

Note. In this theorem (and the proof that follows) neighbourhoods are not assumed to be open. That is, a neighbourhood of a point $x$ is a set $A$ such that $x$ lies in the interior of $A$ .

Proof of theorem. Let $X$ be a topological space in which every point has a closed Hausdorff neighbourhood. Suppose $a,b\in X$ are distinct. It suffices to show that $a$ and $b$ have disjoint neighbourhoods. By assumption, there is a closed Hausdorff neighbourhood $N$ of $b$ . If $a\notin N$ , then $X\setminus N$ and $N$ are disjoint neighbourhoods of $a$ and $b$ (as $N$ is closed).

So suppose $a\in N$ . As $N$ is Hausdorff, there are disjoint sets $U_0,V_0\subseteq N$ that are open in $N$ , such that $a\in U_0$ and $b\in V_0$ . There are open sets $U$ and $V$ of $X$ such that $U_0=U\cap N$ and $V_0=V\cap N$ . Note that $U$ is a neighbourhood of $a$ , and $V$ is a neighbourhood of $b$ . As $N$ is a neighbourhood of $b$ , it follows that $V\cap N$ (that is, $V_0$ ) is a neighbourhood of $b$ . We have $U\cap V_0=U_0\cap V_0=\emptyset$ . So $U$ and $V_0$ are disjoint neighbourhoods of $a$ and $b$ . QED.