Compact Hausdorff Space with no Isolated Points is Uncountable/Lemma

Lemma for Compact Hausdorff Space with no Isolated Points is Uncountable
Let $T = \left({S, \tau}\right)$ be a Hausdorff space with no isolated points.

Let $U \subseteq S$ be a non-empty open set in $T$.

Let $x \in S$.

Then there is a non-empty open subset $V \subseteq U$ such that $x \notin V^-$, where $V^-$ is the closure of $V$.

Proof
First we show that there is a $y \in U$ such that $y \ne x$:

By Law of Excluded Middle, either $x \in U$ or $x \notin U$.

Let $x \in U$.

We have that $x$ is not an isolated point.

Therefore:
 * $\exists y \in U: y \ne x$

Let $x \notin U$.

As $U$ is non-empty it has an element $y$, and $y \ne x$.

Thus in either case, there is a $y \in U$ such that $y \ne x$.

By Closure Condition for Hausdorff Space, there is an open set $V$ such that $y \in V$ and $x \notin V^-$.