Discrete Space is Fully T4

Theorem
Let $T = \struct {S, \tau}$ be a discrete topological space.

Then $T$ is fully $T_4$.

Proof
Consider the set $\CC$ of all singleton subsets of $S$:
 * $\CC := \set {\set x: x \in S}$

From Discrete Space has Open Locally Finite Cover:
 * $\CC$ is an open cover of $T$ which is locally finite.

Let $x \in S$.

The star of $x$ with respect to $\CC$ is defined as:
 * $\displaystyle x^* := \bigcup \set {U \in \CC: x \in U}$

That is, the union of all sets in $\CC$ which contain $x$.

But here the only set of $\CC$ containing $x$ is $\set x$.

Now let $\VV$ be an open cover of $T$.

From the above, it follows that $\CC$ is a cover for $S$ such that:
 * $\forall x \in S: \exists U \in \VV: x^* \subseteq U$

where $x^* = \set x$ is the star of $x$ with respect to $\CC$.

That is, $\CC$ is a star refinement of $\VV$.

Therefore, by definition, $T$ is fully $T_4$.