Neighborhood of Point in Metrizable Space contains Closed Neighborhood

Theorem
Let $T = \left({S, \tau}\right)$ be a metrizable topological space.

Let $x \in S$ be an arbitrary point of $T$.

Let $N$ be a neighborhood of $x$.

Then $N$ has as a subset a neighborhood $V$ of $x$ such that $V$ is closed.

Proof
Since $N$ is a neighborhood of $x$, $\exists$ an open set $U \subseteq N$ containing $x$ by definition.

As $(S, \tau)$ is metrizable, the set $U$ is open with respect to some metric space $(S, d)$, hence $\exists \epsilon > 0$ such that the open $\epsilon$-ball of $x$, $\mathcal{B}_{\epsilon}(x)$, is a subset of $U$.

Taking $\zeta = \frac{1}{2}\epsilon$, we have that the closed $\zeta$-ball of $x$ is contained in $\mathcal{B}_{\epsilon}(x)$ and hence is a subset of $U \subseteq N$.

This closed ball is a closed neighborhood of $x$ in $(S, \tau)$, and is a subset of $N$, so we are done.