If each point in a Hausdorff space $X$ has a nbd whose closure is regular then $X$ itself is regular

here's a problem from Dugundji's book:

Let $X$ be Hausdorff and assume that each $x \in X$ has a nbd $V$ such that $\overline{V}$ is regular. Prove $X$ is regular.

Recall that $X$ is regular iff every $x \in X$ has a local base of closed nbds. So here's what I tried:

Let $x$ be in $X$ and let $W$ be any open set containing $x$. By assumption $x$ has a nbd V such that $\overline{V}$ is regular. This means $\overline{V}$ has a base consisting of closed neighborhoods.

Note $W \cap \overline{V}$ is open in $\overline{V}$ so we may find a closed set $C$ and an open set $U$ such that:

$x \in U \subset C \subset W \cap \overline{V}$.

Since $C$ is closed then $\overline{U} \subset C$.

Hence $x \in U \subset C \subset \overline{W} \subset W \cap \overline{V} \subset W$.

Thus $X$ is regular. Is this OK, if not, how to prove it? thanks in advance!

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Even without looking too closely at the details, there are three immediate problems with your argument. First, your set $U$ is only guaranteed to be open in $\text{cl }V$, not in $X$. Secondly, you never use the observation that $\text{cl }U \subseteq C$. And finally, your penultimate line implies that $\text{cl }W \subseteq W$ and hence that $W$ is closed, which clearly need not be the case for an arbitrary open nbhd of $x$.

It's easier to use Dugundji's definition of regularity or one of the equivalent statements that immediately follow it. Let $F$ be a closed subset of $X$ and $x \in X \setminus F$; we'll find an open nbhd of $x$ whose closure is disjoint from $F$. By hypothesis there is an open set $V$ such that $x \in V$ and $\text{cl }V$ is regular. Let $W = V \setminus F$; $W$ is an open nbhd of $x$ in the regular space $\text{cl }V$, so there is a set $U$ such that $x \in U \subseteq \text{cl }U \subseteq W$ and $U$ is open in $\text{cl }V$. Since $U$ is open in $\text{cl }V$, there is an open set $U_0$ in $X$ such that $U = U_0 \cap \text{cl }V$. Let $U_1 = V \cap U_0$; $V$ is open in $X$, so $U_1$ is as well, and clearly $x \in U_1 \subseteq \text{cl }U_1 \subseteq \text{cl }U \subseteq W \subseteq X \setminus F$.

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