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.