Proof idea - closed subspace of compact metric space is compact I came up with the following proof and I am afraid it is too simple and that I missed something...
Question: If $X$ is a compact metric space and $M \subset X$ is closed, show $M$ is compact.
Proof idea: We know that $X$ compact gives that $X$ is closed and bounded. To show that $M$ is compact we must just show that $M$ is bounded since it is given that $M$ is closed. But since $M \subset X$ and $X$ is bounded doesn't that immediately give that $M$ is bounded? 
 A: If $X = \mathbb{R}^{n}$, then compactness is equivalent to closedness and boundedness. Otherwise, the equivalence need not hold. 
A simple argument to see the desired result is as follows. Note that there is some open cover $\{ U_{\alpha} \}$ of $M$. So $\{ U_{\alpha} \} \cup F^{c}$ is an open cover of $X$. Note that there is some finite subcover of the cover. 
A: The general definition of a subset of the ground set in a topological space being compact is that any open cover has a finite subcover. In general metric spaces, closed and bounded is not equivalent to being compact, though this holds in $\mathbb{R}^n$. 
In order to proceed, suppose $X$ is a compact topological space with $M$ any closed subset with some arbitrary  open cover $C = \{O_i\}_{i\in I}$ of $M$. Then $N:=X\setminus M$ is open and adding it into the cover to form $C':=C\cup \{N\}$ is an open cover of all of $X$. Thus there is a finite subcover of $C'$ which covers $X$. Throwing out $N$ if it is included in the finite subcover of $C'$, we arrive at a finite subcover of $C$ covering $M$ as desired.  
