Discrete Space is Paracompact

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

Then $T$ is paracompact.

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 which is locally finite.

Consider any open cover $\mathcal V$ of $S$.

By definition, $\forall x \in S: \exists V \in \mathcal V: x \in V$.

That is, $\forall x \in S: \exists V \in \mathcal V: \left\{{x}\right\} \subseteq V$.

That is, every element of $\mathcal C$ is contained in some element of $\mathcal V$.

Thus by definition, $\mathcal C$ is an open refinement of $\mathcal V$ which is (as we saw earlier) locally finite.

So $T$ is paracompact, by definition.