Suppose that $X$ is a compact Hausdorff space and let $\mathcal{F}\subseteq\mathcal{P}(X)$ be a family of closed sets in $X$ whit the FIP property. Let $U\subseteq X$ be an open set such that $\bigcap\mathcal{F}\subseteq U$. I must to show that there exists $F_0,\dots,F_n\in\mathcal{F}$ such that $\cap_{i\leq n}F_i\subseteq U$.
Proof: Consider $\mathcal{F}^c:=\{F^c:F\in\mathcal{F}\}$. Note that $\mathcal{F}^c\cup\{U\}$ is an open covering of $X$ and, as $X$ is compact, there exists $F_0,\dots,F_n\in\mathcal{F}$ such that $X=\cup_{i\leq n}F_i^c\cup U$. Then $\cap_{i\leq n}F_i\subseteq U$.
Question: In my proof I never use that $X$ is a Hausdorff space so, Is that hypotesis really necessary? Also, I feel that the condition that $\mathcal{F}$ has the FIP property is not necessary and it can be changed by other e.g, each element of $\mathcal{F}$ is not empty closed set. Im I right?
By definition all $F \in \mathcal{F}$ are non-empty (1 set intersections are also finite intersections) and a standard equivalence of compactness (which holds regardless of separation axioms) allows us to conclude that $\bigcap \mathcal{F}$ is non-empty (this is where the FIP plays an important role).
The proof you gave is fine and does not need Hausdorffness. Hausdorffness can be omitted.
If $\mathcal{F}$ does not have the FIP, there are $F_1,\ldots F_n \in \mathcal{F}$ with empty intersection, and $\bigcap \mathcal{F}=\emptyset$ too. So for any $U$ (open or not) we can just take those $F_1,\ldots F_n$. So the whole statement trivialises and becomes meaningless for non-FIP families.