Proof regarding nested compact subsets of a metric space Let $\left\lbrace X_n \right\rbrace$ be a sequence of compact subsets of a metric space $M$ with $X_1 \supset X_2 \supset X_3 \supset \dots$. Prove that if $U$ is an open set containing $\cap X_n$, then there exists $X_n \subset U$.
This proof is tricky for me because I can't use many facts about compactness beyond its definition (a metric space $M$ is compact if every open cover of $M$ has a finite subcover). For example, I can't use the fact that every sequence in a compact metric space has a convergent subsequence, or the fact that a compact subset is closed and bounded. I do "know" that a finite union of compact subsets is compact and the intersection of compact subsets is compact.
My main idea has been a proof by contradiction in which I assume that $X_n \not\subset U$ for all $n$. Then I find a point in each $X_n$ that is not in $U$ and build a sequence $x_n$ that converges to a point in $U$. Since $x_n$ is in the closed subset $M \backslash U$, I would have a contradiction. But I can't figure out how to show that such a sequence exists.
 A: Given the tools available to you, sequences aren’t the best way to go.
HINT: Let $K=X_1\setminus U$; $K$ is a closed subset of the compact set $X_1$, so $K$ is compact. For each $n\in\Bbb Z^+$ let $V_n=M\setminus X_n$, and let $\mathscr{V}=\{V_n:n\in\Bbb Z^+\}$.


*

*Show that $\mathscr{V}$ is an open cover of $K$.  

*Conclude that there is an $n\in\Bbb Z^+$ such that $K\subseteq V_n$.  

*What does this tell you about $X_n$?

A: Suppose not. For each $n$ let $p_n\in X_n$ \ $U.$ Then $(p_n)_{n\in N}$ is a sequence in $X_1$ so there exists $q\in X_1$ and a strictly increasing $f:N\to N$ such that $(p_{f(n)})_{n\in N}$ converges to $q.$
And for $j>1, $ the sequence $(p_{f(n)})_{f(n)\geq j}$ is a sequence in $X_j,$ also converging to $q,$ so $q\in X_j.$ 
So $q\in \cap_{n\in N}X_n.$ So $q\in U.$
Now $(p_{f(n})_{n \in N}$ converges to $q ,$and $f$ is strictly increasing, so any nbhd of $q$ contains $p_{f(n)}$ for all but finitely many $n.$ And $U$ is a nbhd of $q$ (because $q\in U$).  So $p_{f(n)}\in U$ for all but finitely many $n .$
This contradicts $\forall n\;(p_{f(n)}\not \in U).$
