Discrete Space is Fully T4

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

Then $T$ is fully $T_4$.

Proof
Consider the set:
 * $\mathcal C := \left\{{\left\{{x}\right\}: x \in S}\right\}$

That is, the set of all singleton subsets of $S$.

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

Let $x \in S$.

The star of $x$ with respect to $\mathcal C$ is defined as:
 * $\displaystyle x^* := \bigcup \left\{{U \in \mathcal C: x \in U}\right\}$

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

But here the only set of $\mathcal C$ containing $x$ is $\left\{{x}\right\}$.

Now let $\mathcal V$ be an open cover of $T$.

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

where $x^* = \left\{{x}\right\}$ is the star of $x$ with respect to $\mathcal C$.

That is, $\mathcal C$ is a star refinement of $\mathcal V$.

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