Can you please let me know if my "proof" is correct: ?
Let $Y$ be a Hausdorff space and assume that each $y \in Y$ has a neighborhood $V$ such that $\overline{V}$ is regular. Prove $Y$ is regular.
Proof:
In order to show that $Y$ is regular we are going to show that for each $y \in Y$ and each open set $U$ which contains $y$ there exists an open neighborhood $W$ of $y$ such that $y \in W \subseteq \overline{W} \subseteq U$.
First observation: since regularity is hereditary then $V$ is regular.
Now let $y \in Y$ and $U$ a neighborhood of $y$. By assumption there exists a neighborhood $V$ of $y$ such that $\overline{V}$ is regular and hence $V$ is regular. Observe $U \cap V$ is a non-empty subset of $V$ and this set is open in $V$ as well. Since $V$ is regular we can find a non-empty subset $S$ of $V$ such that:
$S \subseteq \overline{S} \subseteq U \cap V \subseteq U$
But since $V$ is open then $S$ is open in $X$ so the same set $S$ given by regularity of $V$ shows that $U$ is regular as well.
Is the above incorrect?
Thank you.