Discrete Space has Open Locally Finite Cover

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

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

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

Then $\mathcal C$ is an open cover of $T$ which is locally finite.

This cover is the finest cover on $S$.

That is, if $\mathcal V$ is a cover of $T$, then $\mathcal C$ is a refinement of $\mathcal V$.

Proof
We have that:
 * $\forall x \in S: \exists \left\{{x}\right\} \in \mathcal C: x \in \left\{{x}\right\}$

and so $\mathcal C$ is a cover for $S$.

Then from Set in Discrete Topology is Clopen, it follows that $\mathcal C$ is an open cover of $T$.

From Point in Discrete Space is Neighborhood, every point $x \in S$ has a neighborhood $\left\{{x}\right\}$.

This neighborhood $\left\{{x}\right\}$ intersects exactly one element of $\mathcal C$, that is: $\left\{{x}\right\}$ itself.

As $1$ is a finite number, the result follows from definition of locally finite.

Now consider any 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 a refinement of $\mathcal V$.