How to use compact property to show this proof from topology

Question

Let X be a compact Hausdorff space, {$F_{n}$ | n $\in$ $\mathbb{N}$,} a descending collection of closed subsets of X; and O an open set containing $\cap$$F_{n}$. Show that $\exists$ N such that $F_{n} $$\subseteq$ O $\forall$ n $\ge$ N.

Ok I have made some progress: - I know all these closed sets are compact as X is compact and any closed set of a compact space is compact (thus every open cover of each closed set has a finite subcover) - I think maybe using the finite intersection property may be useful

I am looking for a way to approach this problem. How to set it up. Thank you

Answer 1

If all sets $F_n\setminus O$ are non-empty, then it contardicts the finite intersection property.

Answer 2

Suppose (by contradiction) that $(n_k)_{k\in \mathbb N}$ is a strictly increasing sequence in $\mathbb N$ such that $F_{n_k}\not \subset O$ for each $k.$ Choose $p_k\in F_{n_k}$ \ $O.$ Let $G_k=\{p_j:j\geq k\}.$ Then $$\emptyset \ne \overline {G_{k+1}}\subset \overline {G_k}\subset \overline {F_{n_k}}=F_{n_k}.$$ Let $G= \cap_{k\in \mathbb N}\overline {G_k}.$ We have $G\ne \emptyset.$ Take any $p\in G$. Then $p$ belongs to $\cap_{m\in \mathbb N}F_m,$ because for any $m$ we may take $n_k>m,$ and $p\in \cap_{j\geq k}\overline {G_{n_j}} \subset F_{n_k}\subset F_m.$ So $p\in O.$ But then $O$ is a nbhd of $p,$ with $p\in \overline {G_k}$ (for any $k$), which implies $G_k\cap O\ne \emptyset,$ a contradiction.