Topologically Distinguishable Points are Distinct

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

Let $x, y \in X$ be topologically distinguishable.

Then the singleton sets $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are disjoint and so:
 * $x \ne y$

Proof
Let $x$ and $y$ be topologically distinguishable.

Then either:
 * $\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$

or:
 * $\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$

Suppose $x = y$.

Then:
 * $\neg \left({\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}\right)$

and:
 * $\neg \left({\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}\right)$

Hence the result by Proof by Contradiction:
 * $x \ne y$

and so by Singleton Equality:
 * $\left\{{x}\right\} \ne \left\{{y}\right\}$