Equivalent criterion of local compactness

Question

I'm trying to understand the proof of the following criterion:

Let $X$ be a Hausdorff's space and locally compact in $x\in X$, then for all $U$ open that contains $x$ exists an open set $V$ such that $x \in \overline{V} \subset U$.

In a part of the proof, they said that if $C$ is a closed subset and $V,W$ disjoint open sets such that $C \subset W$ and $x \in V$ then $\overline{V}$ and $C$ are disjoint sets but I dont understand why. Could anybody explain me this part? :)

Answer 1

Even $W$ is disjoint from $\bar V,$ since $V$ is contained in the closed subset $X\setminus W$ hence so is $\bar V.$

Answer 2

Because if $\bar V\cap C$ were non empty, then for any $t\in \bar V\cap C$, $W$ is a neighborhood of $t$. Since $t\in \bar V$, we must have $W\cap V\ne \emptyset$, which is a contradiction. So $\bar V\cap C=\emptyset.$