Let $G$ be an open subset of a metric space $(X,d)$. Let $K$ be a compact subset of $(X,d)$. Show that $K - G$ is compact. A subset $K$ of a metric space $(X,d)$ is compact if every sequence $\{p_n\}$ in $K$ has a subsequence $\{p_{ni}\}$ which is convergent to a point in $K$.
A subset $G$ of a metric space $(X,d)$ is open if every point in $G$ is an interior point of $G$. There exists a ball of some radius around every point in $G$ which is entirely contained in $G$.
An element is a member of $K-G$ if it is a member of $K$ but not a member of $G$. 
This is all the information I've got. How should I start my proof?
 A: Hints:


*

*Take a sequence $(x_n)_n$ from $K$ \ $ G$

*Use that $K$ is compact

*Use that $G$ is open, hence $X$ \ $G$ is closed

*Conclude
A: This is how you could start. It is only a minor step from where I leave you.
Let $x_n$ be a sequence in $K-G$. It is in particular a sequence in $K$, hence by $K$'s compactness it admits a convergent subsequence $k_{n_i}$ with limit in $K$. 
Claim: This limit lies in $K-G$. This is now what you have to do! Think about what would happen, if this weren't true. 
A: Usually, the definition of compactness is that:

every open cover of the space has a finite subcover.  

The definition of compactness you gave is actually called sequential compactness.  
However, a metric space is compact if and only if it is sequentially compact, so we can use the definition of sequential compactness to prove compactness, as you suggested.
Now, let $(X, d)$ be a metric space.  Suppose $K \subseteq X$ is compact, and $G \subseteq X$ is open.  We want to show $K - G$ is compact.  Let's do it with cases:
1) If $K \cap G = \emptyset$, then $K - G = K$, so $K - G$ is compact, since $K$ is compact.
2) If $K \cap G \neq \emptyset$, then:
a) if $K \subseteq G$, then $K - G = \emptyset$, and $\emptyset$ is always compact (why?),
b) if $K \not \subseteq G$, then $K - G \neq \emptyset$, so let $\{ x _{n} \}$ be a sequence in $K - G$.  We want to show there is a subsequence of $x_{n}$ which converges in $K - G$.  Suppose not.  Then for every $x \in K - G$ not in $\{x_{n}\}$, there is some $\epsilon_{x} > 0 $ such that $B(x, \epsilon_{x})$ contains no elements of the sequence $x_{n}$ (why?).  Also, for each $m$, $\exists \epsilon_{m} > 0$ such that $B(x_{m}, \epsilon_{m}) \cap \{ x_{n} \}_{n = 1}^{\infty} = x_{m}$.  
Now, take $\left (\bigcup \limits_{x \in K - G} B(x, \epsilon_{x}) \right ) \cup \left ( \bigcup \limits_{n = 1}^{\infty} B(x_{n}, \epsilon_{n}) \right ) \cup G$.  This is an open cover of $K$ with no finite subcover (why?), which means $K$ is not compact !  This is a contradiction since we assumed $K$ was compact.  Thus, $K - G$ is compact.
So, in every case, we showed that if $K \subseteq X$ is compact, and $G \subseteq X$ is open, then $K - G$ is compact, as desired.
A: Exactly so: I think the approach suggested by the question would be to let $S$ be the set $K - G$ and let {$s_n$} be a sequence of points in S. Since points in S are points in K there must be a convergent subsequence {$s_{ni}$} which converges to a point in K. Since S are those points in K but not G it suffices to show that there is indeed a sub-sequence which converges to a point in K which is also not in G. This convergent sub-suquence and its limit point are probably the best place to start.
