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?

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!
