Let $\displaystyle P(N)$ be the power set of $\displaystyle N$.

We show that given $\displaystyle A \in P(N)$, either $\displaystyle A \in F$ or $\displaystyle N-A \in F$. Indeed, both $\displaystyle A$ and $\displaystyle N-A$ cannot be in $\displaystyle F$ because their intersection is empty. Thus the map $\displaystyle g : F \rightarrow (P(N)-F) $ which maps $\displaystyle S \mapsto (N-S)$ is well-defined, and a bijection because $\displaystyle F$ contains $\displaystyle 2^{n-1}$ elements. Since the domain and the image of this map partition $\displaystyle P(N)$, one of $\displaystyle A$ or $\displaystyle N-A$ must lie in $\displaystyle F$.

Now suppose $\displaystyle F$ is not closed under intersection. Then there exists $\displaystyle A,B \in F$ such that $\displaystyle (A \cap B) \notin F$. By the above, we must have that $\displaystyle N-(A \cap B)$ is in $\displaystyle F$. But $\displaystyle (N-(A \cap B))\cap A \cap B = \emptyset$, which contradicts the hypothesis that any three elements of $\displaystyle F$ have a nonempty intersection. Therefore $\displaystyle F$ is closed under intersection, and $\displaystyle \left(\cap_{A \in F}A\right)\in F$; since $\displaystyle F$ does not contain $\displaystyle \{\emptyset\}$ we are done.