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 set of open sets of $X$.

Then $\mathcal F$ is countable.

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

From Open Set Characterization of Denseness, each element of $\mathcal F$ contains an element of $S$.

From Union of Set of Sets with Set Intersecting with All:


 * $\mathcal F = \displaystyle \bigcup_{x \mathop \in S} \left\{{V \in \mathcal F: x \in V}\right\}$

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