First, either you’ve miscopied the problem, or the person who wrote it misused some of the notation rather badly. If $\Bbb F$ is the set of all closed subsets of the closed set $F$, then every member of $\Bbb F$ is a subset of $F$; $\Bbb F$ does not contain any collections of closed sets. $\tilde F$ is actually the set of all collections $\{F_\alpha:\alpha\in A\}\subseteq\Bbb F$ (note the subset relation, not the membership relation) satisfying a certain condition. That condition says that the collection is centred, or has the finite intersection property: $\{F_\alpha:\alpha\in A\}\in\tilde F$ if and only if
- $\{F_\alpha:\alpha\in A\}\subseteq\Bbb F$, and
- for every finite $\{\alpha_1,\dots,\alpha_n\}\subseteq A$, $\bigcup\limits_{k=1}^nF_{\alpha_k}\ne\varnothing$.
Your problem is to prove that $F$ is compact iff $\bigcap_{\alpha\in A}F_\alpha\ne\varnothing$ for every family $\{F_\alpha:\alpha\in A\}\in\tilde F$. In other words, $F$ is compact if and only if every centred family of closed subsets of $F$ has non-empty intersection.
This of course requires that you show two things: if $F$ is compact, then every centred family of closed subsets of $F$ has non-empty intersection, and if every centred family of closed subsets of $F$ has non-empty intersection, then $F$ is compact. The same basic observations are used to prove both directions:
Let $\mathscr{U}$ be a family of open subsets of $F$, and let $\mathscr{F}=\{F\setminus U:U\in\mathscr{U}\}$.
- $\mathscr{U}$ covers $F$ if and only if $\bigcap\mathscr{F}=\varnothing$.
- If $\mathscr{U}$ covers $F$, $\mathscr{U}$ has a finite subcover if and only if $\mathscr{F}$ is not centred (or if you prefer, is not in $\tilde F$).
Prove these, and you’re well on your way.