How can I prove this statement about filters?

Yes, this is true and your proof valid. A filter $\mathcal F$ on a finite set $X$ is a subset of $\mathscr{P}(X)$ and hence is finite, i.e. $\mathcal{F}$ is a finite set of subsets. One axioms of filters tells us that $\mathcal{F}$ is closed under binary intersection, and indeed a standard induction argument (which is often skipped as too self-evident) shows that in fact $F_0:=\mathcal{F}$ is closed under all finite intersections, and in particular $\bigcap \mathcal{F} \in \mathcal F$. As $F_0 \neq \emptyset$ (a filter does not contain $\emptyset$ by another axiom), this means that $\mathcal F$ is a fixed filter (i.e. has non-empty intersection).

Using the enlargement axiom it is in fact easy to see that then the following holds:

$$\mathcal F = \{A \subseteq X\mid F_0 \subseteq A\}$$