Singleton Set in Discrete Space is Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space where $\tau$ is the discrete topology on $S$.

Let $x \in S$.

Then $\left\{{x}\right\}$ is compact.

Proof 1
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\left\{{x}\right\}$.

Then from Interior Equals Closure of Subset of Discrete Space we have that $\left\{{x}\right\}$ equals its closure.

As $\left\{{\left\{{x}\right\}}\right\}$ is (trivially) an open cover of $\left\{{x}\right\}$, it follows by definition that $\left\{{x}\right\}$ is compact.

Proof 2
Follows directly from Finite Topological Space is Compact.