Let $(X, \mathcal{T})$ be a Hausdorff topological space.
Definition. $X$ has the finite intersection property (FIP), if for every family of closed subsets $(A_i)_{i \in I} \subset X$ with $\bigcap_{i \in I} A_i = \emptyset$ the exist indices $i_1, \ldots, i_k$, so that already $\bigcap_{\ell = 1}^{k} A_{i_{\ell}} = \emptyset$.
I've already proven that $X$ is compact iff $X$ has the FIP, which basically follows from deMorgans Principles and the fact that $(A_i^C)_{i \in I}$ are an open cover of $X$.
Let $(A_i)_{i \in I} \subset X$ be a compact family. Prove that if $\bigcap_{i \in I} A_i = \emptyset$ the existence of indices $i_1, \ldots, i_k$, so that $\bigcap_{\ell = 1}^{k} A_{i_{\ell}} = \emptyset$ already follows.
My ideas
I know that finite unions of compact sets and countable intersections are compact because $(X, \mathcal{T})$ is Hausdorff, but haven't been able to use that.
I also tried reaching a contradiction by assuming that for all finite sets of inidices $\{i_1, \ldots, i_k\}$ the intersection $\bigcap_{\ell = 1}^{k} A_{i_{\ell}}$ is nonempty but I haven't been able to show that then the infinite intersection is nonempty, which would by the desired contradiction.
All help is greatly appreciated.
Edit (my approach using the hint)
And $A_{i_0} \subset X$ is a compact subspace, so we have a family $(\tilde{A}_i := A_{i_0} \cap A_i)_{i \in I}$ in a compact space with $\bigcap_{i \in I} \tilde{A}_i = \emptyset$ and by FIP we know that finitely many indices are enough, so already $\bigcap_{\ell = 1}^{k} \tilde{A_{i_{\ell}}} = \emptyset$. Now we have for those finitely many indices we have $$ \emptyset = \bigcap_{\ell = 1}^{k} A_{i_0} \cap A_{i_{\ell}} = A_{i_0} \cap \bigcap_{\ell = 1}^{k} A_{i_{\ell}} $$ And since $A_{i_0} \neq \emptyset$, we have $\bigcap_{\ell = 1}^{k} A_{i_{\ell}}$ as required.
Given some family of compact subsets $A_i, i \in I$ with empty intersection:
Fix some $A_{i_0}$ (which is compact), and define $B_i = A_{i_0} \cap A_i$.
Then some $B_i$ can be empty (and we are done, using the finite set $\{i, i_0\}$) or all $B_i$ are non-empty and $\bigcap_{i \in I} B_i = \bigcap_{i \in I} A_i =\emptyset$ and the already proved fact for the compact space $A_{i_0}$ (we use Hausdorffness in that all $B_i$ are closed in $A_{i_0}$ so that that theorem applies) implies that for a finite subset $J$ of $I$ we have that $\bigcap_{i \in J} B_i = \emptyset$ and then we have that for the finite subset $J'=J \cup \{i_0\}$ of $I$ that $$\bigcap_{i \in J'} A_i = \emptyset$$
as required.