Compact Hausdorff Space with no Isolated Points is Uncountable/Lemma

Theorem
Let $(X, \tau)$ be a Hausdorff space with no isolated points.

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

Let $x \in X$.

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 $x \notin U$.

If $x \in U$, then since $x$ is not an isolated point, there must be a $y \in U$ such that $y \ne x$.

If $x \notin U$, then since $U$ is non-empty it has an element $y$, and $y ≠ x$.

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 $x \notin V^-$.