If you apply the definition of “regular” to the point $x$ and the closed set $U^{c}$ you get disjoint open sets $V \ni x$ and $W \supset U^c$. It follows that $\overline{V}$ is contained in the closed set $W^c \subset U$.
Added:
It may be worth noting that the following properties of a topological space $X$ are equivalent:
- For every point $x$ and every open set $U \ni x$ there is an open neighborhood $V$ of $x$ with $\overline{V} \subset U$.
- The space is regular in the sense that for every point $x$ and every closed set $C$ with $x \notin C$ there are disjoint open sets $U \ni x$ and $V \supset C$.
1. implies 2.:
If $x \in X$ and $C \subset X$ is closed with $x \notin C$ put $U = C^{c}$. Then $x \in U$ and $U$ is open. By 1., we can find an open neighborhood $V$ of $x$ with $\overline{V} \subset U$. Then $W = \overline{V}^c \supset C$ and $V \cap W = \emptyset$ are disjoint open sets separating $x$ and $C$.
2. implies 1.:
That's the argument given at the beginning of this answer.