text: Q:
  
  If $X$ is a compact metric space and $E_n$ is closed nonempty subset, show that $\cap_{n=1}^\infty E_n$ is nonempty.
  
  Suppose that $(X,d)$ is a compact metric space and $(E_n)$ is any sequence of nonempty closed subsets of $X$ with $E_{n+1}\subset E_n$ for all $n\in\mathbb{N}$. Show that $\cap_{n=1}^\infty E_n$ is nonempty.
  What I know: Since $X$ is compact, we need to use the fact that every sequence in $X$ has subsequence which converges in $X$. $E_n$ is nonempty and closed, that means it contains all its limit points. And $E_n$ is also compact right? 
  My doubts are: How can we construct a sequence that has a subsequence that has a limit in $X$? I don't think we should use prove by contradiction here. My guess is to show the limit that we have from the fact that $X$ is compact is in the intersection. But I don't even have the sequence yet. 
  Can anyone please give some clue on the line of reasoning of the proof? Many thanks!
  
  A:
  
  Alternative idea:
  $$\bigcap_{n=1}^\infty E_n = \emptyset\implies \bigcup_{n=1}^\infty(X\setminus E_n) = X$$
  
  A:
  
  For each $n$ choose some $x_n \in E_n$ to get a sequence $\{ x_n \}$. By passing to a subsequence if necessary we may assume that $x_n \to x$ for some $x \in X$. Now for any $n$ all but the first $n$ terms (at most) of this sequence lie in $E_n$. If we ignore these first few terms we still have a sequence in $E_n$ converging to $x$. Since $E_n$ is closed, $x \in E_n$. This holds for any $n$, so $x \in \cap_{n=1}^{\infty} E_n$.
  
  