I dont understand the following proof:
Lemma. Let $X$ be normal, $C$ closed, and $U$ open with $C \subseteq U$. Then there exists an open set $V$ such that $C \subseteq V \subseteq \operatorname{cl}(V ) \subseteq U$.
Proof. Since $C$ and $U^\complement $ are disjoint closed sets, by normality there exist disjoint open sets $V\supseteq C$ and $W \supseteq U^\complement $. If $x\in\operatorname{cl}(V )\cap U^\complement $, then $W$ , as a neighborhood of $x$, meets $V$, a contradiction. Therefore $\operatorname{cl}(V)\subseteq W$.$\Box$
I guess that the expression "$W$ meets $V$" means that $V\cap W\neq \emptyset $, however I cannot see why. I mean: if $W$ is a neighborhood of $x$ and $x\in \partial V$ then I dont see why necessarily $V\cap W\neq \emptyset $. Can someone enlighten me, please?
Because we are assuming that $x\in\overline V$. Therefore, by the definition of closure, every neighborhood $W$ of $x$ meets $V$ (that is, $W\cap V\neq\emptyset$).