Proposition 6

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

Let $x \in X$ and let $V$ be a closed neighbourhood of $x$ in $X$ such that the subspace $V$ is Hausdorff. Then the closed neighbourhoods of $x$ in $V$ have $\{x\}$ as their intersection (axiom (H\textsuperscript{'})); but they are also closed neighbourhoods of $x$ in $X$ (§ 3, no. 1) and therefore $X$ satisfies (H\textsuperscript{i}).

There exist non-Hausdorff spaces in which every point has a Hausdorff neighbourhood (Exercise 7).