Proving that $(X,\tau)$ is compact iff every filter has a finer filter that converges I'm having some trouble understanding one step in a proof. The objective is to prove the following:

Let $(X,\tau)$ be a topological space. Then $(X,\tau)$ is compact if and only if for every filter $\cal F$ on $(X,\tau)$ there is a filter $\cal F_1$ which is finer than $\cal F$ and converges.

I'm having some trouble with the following part of the proof:
Recall that a topological space is compact if and
only if for every family $\cal S$ of closed subsets with the finite intersection property,$$\bigcap_{S_i\in\mathcal S} S_i \neq \emptyset$$
[...]
Assume that for every filter on $(X, τ)$ there is a filter that is finer
than it which converges. Let $\cal S$ be a family of closed subsets of $(X, τ)$ with the finite intersection property. So there is a filter $\cal F$ on $(X, τ)$ which contains $\cal S$. So by assumption, there exists a point $x_0$ in $X$ and a filter $\mathcal F_1 ⊇ \mathcal F$ such that $\mathcal F_1 → x_0$. So we can conclude that $\mathcal F_1 ⊇ \mathcal N_{x_0}$, the neighborhood filter
of $x_0$. Thus for each $N_{x_0} \in \mathcal N_{x_0}$, $N_{x_0} ∩ F \neq \emptyset$, for each $F ∈ \mathcal F_1$. In particular, $N_{x_0} ∩ S_i \neq \emptyset$, for every $S_i ∈ \cal S$. $\underline{\text{this now is the part I'm having some trouble with:}}$
Therefore $$x_0 ∈ \bigcap_{S_i \in \mathcal S} S_i$$
So $(X, τ)$ is compact.

How can we conclude from "$N_{x_0} ∩ S_i \neq \emptyset$, for every $S_i ∈ \cal S$" that $x_0 ∈ \bigcap_{S_i \in \mathcal S} S_i$?
Because $N_{x_0} ∩ S_i \neq \emptyset$ doesn't nesesseraly imply that $x_0 \in S_i$.

This uses the fact that $S_i$ is closed.  You know $S_i$ intersects every neighborhood of $x_0$, and so $x_0$ is in the closure of $S_i$.  Since $S_i$ is closed, this tells you that actually $x_0\in S_i$.
