Show that $K_1\cap K_2\cap \dots,K_N$ is compact Let $K_1,K_2,\dots,K_N$ be compact subsets of the mectric space $(X,d)$. 
Show that $K_1\cap K_2\cap \dots,K_N$ is compact.
My approach:
I think I should use the definition of compact sets in my textbook:
Let $(X,d)$ be a metric space. A subset $K\subseteq X$ is compact if every sequence in $K$ has a convergent subsequence with limit in $K$.
I can't get further than this. Can you help?
Update:
What if it was union instead of intersection?
 A: Observe that in a metric space compact sets are closed. Intersection of closed sets are closed. And closed subset of a compact set is compact. These three facts imply the conclusion.
These all statements are valid if we consider a Hausdorff topological space, as a generalisation of metric space. So the above claim is true for an Hausdorff space.
A: Take an arbitrary sequence $\{x_n\}\subset K_1\cap K_2\cap...\cap K_N$. This means that the sequence is contained in each $K_i,\,i=1,..,N$. Now for $K_1$ you can find a convergent subsequence (still contained in all $K_i$ and in particular in $K_2$ ) with limit in $K_1$. From it you find a further convergent subsequence in $K_2$ with limit also in $K_2$, and so on. Finally you find a convergent subsequence which has a limit in each $K_i$ as $K_i$ are compact. Also the limit is unique and is contained in each $K_i\Rightarrow$ is contained in the intersection. 
If you had to prove that $\cup K_i$ is compact, then again taking an arbitrary sequence $\{x_n\}\subset K_1\cup K_2\cup...\cup K_N$ you want to find a convergent subsequence with limit again in the union. You have that each $x_n$ is contained in at least one $K_i$. Notice that because there are finite number of $K_i$, there must be some index $i_0$ such that $K_{i_0}$ contains infinitly many terms of the seuqence, otherwise if each $K_i$ contains only finitly many terms, the sequence would consist only of finite number of terms. So, in $K_{i_0}$ there is a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, and because $K_{i_0}$ is compact there is a convergent (further) subsequence of it with limit in $K_{i_0}$ and therefore also in the union.
A: A set is compact if every cover has a finite subcover.
So if you take the intersection of non-empty sets, each of which has a finite subcover, does it follow the elements in all of them also has a finite subcover? 
A: Actually, arbitrary intersection (not only finite) of compact subsets is also compact. This is pretty easy to see: in metric spaces, compact subsets are closed, so their intersection is also closed. On the other hand, the closed subset of a compact set is also compact: suppose $K$ is compact and $N \subset K$ is closed. Take any open cover $N \subset \bigcup U_i$ of $N$. As $K - N$ is open, consider open cover $K \subset (K - N) \cup \bigcup U_i$. As $K$ is compact, it has some finite subcover $K \subset V_1 \cup ... \cup V_n$. If $K - N$ is one of the $V_i$, say $V_1$, $V_2 \cup ... \cup V_n$ covers $N$, and is a finite subcover of the original cover $N \subset \bigcup U_i$.
Thus, as closed subset of compact set is compact, and intersection of compact set is a closed subset of any of them, it is compact.
On the other hand, a finite union of compact sets is compact, but infinite one no longer necessarily is!
