Point Finite Set of Open Sets in Separable Space is Countable

Theorem
Let $(X, \tau)$ be a separable space.

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

Then $\mathcal F$ is countable.

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

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

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

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