Singleton Set in Discrete Space is Compact/Proof 1

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

Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure.

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