Point Finite Set of Open Sets in Separable Space is Countable

Theorem
Let $\left({X, \tau}\right)$ be a separable space.

Let $\mathcal F$ be a point-finite open cover of $X$.

Then $\mathcal F$ is countable.

Proof
Since $\left({X, \tau}\right)$ is separable, $X$ has a countable dense subset $S$.

Assume WLOG that $\varnothing \notin \mathcal F$.

By the definition of point-finite, $\left\{{V \in \mathcal F: x \in V}\right\}$ is finite for each $x \in S$.

Since $S$ is dense in $X$, $\mathcal F = \displaystyle \bigcup \left\{{V \in \mathcal F: x \in V }\right\}$.

Thus by Countable Union of Finite Sets is Countable, $\mathcal F$ is countable.