Help completing a proof regarding an infinite family of nested closed subsets in a compact metric space

Question

I am trying to prove the following:

"Let $U$ be an open subset in the compact metric space $X$. If $\{S_k : k\in\mathbb{N}\}$ is a collection of closed subsets of $X$ such that $S_{k+1}\subset S_k$ for all $k\in\mathbb{N}$, and $\bigcap_{k=1}^{\infty} S_k \subset U$, then $S_n\subset U$ for some $n\in\mathbb{N}$."

Here is what I have so far:

Being closed subsets of a compact set, each $S_k$ is compact.

Additionally, I have a theorem which states that an infinite collection of nested compact subsets of $X$ has a nonempty intersection. Therefore, $\bigcap_{k=1}^{\infty} S_k \subset U$ implies that there exists an element $p$ such that $p\in\bigcap_{k=1}^{\infty} S_k \subset U$, i.e., each $S_n$ contains at least one point which is an element of $U$. Being open, there exists some $\epsilon >0$ such that $B(p,\epsilon)\subset U$ (I'm not sure if that is important).

I haven't been able to get any further. This problem is given not long after first being introduced to compact sets, so I don't expect the completion of the proof will require much theory. Any hints or a solution or a point in the right direction would be greatly appreciated. Thank you.

Also, I am sorry if this is a duplicate, I haven't been able to find anything like it. I am also not sure if the (analysis) tag is appropriate, I don't know of a more specific tag which is relevant.

Answer 1

Here is a different approach using the definition of compactness:

Note that $U,S_1^c,S_2^c,...$ is an open cover of the compact set $S_1$ (show this!).

Hence there is a finite sub cover, and since the $S_k$ are nested, we have $S_1 \subset U \cup S_n^c$ for some $n$.

Since $S_n \subset S_1$, we have $S_n \subset U \cup S_n^c$, from which it follows that $S_n \subset U$.

Answer 2

Suppose not. Then for all $n$, there exists an $x_n\in S_n$, and not in $U$, ie $x_n\in F={}^c U$. Now $x_n\in S_1$ for all $n$. As $S_1$ is compact, there exists a sub-sequence $x_{n_k}$ of $x_n$ that is convergent to some $x$. As $F$ is closed, we have $x\in F$. Now fix $m$. If $n_k\geq m$, then $x_{n_k}\in S_m$. As $S_m$ is closed, this imply $x\in S_m$, and $x$ is in the intersection of the $S_m$, hence in $U$, contradiction.