How does one show that $K_1 \cap K_2 $ is compact, when $K_1 , K_2$ are compact? Let $(X,d_X) $ be a metric space and $K_1 , K_2 $ be compact subspaces of $X$. Question: how does one show that $K = K_1 \cap K_2$ is compact? 
I tried proving this by noting the following theorem: Let $Y \subseteq X$.
If $X$ is compact and $Y$ is closed in $X$, then $Y$ is compact. 
We know, that compact spaces are closed. So in particular, we know that $K_1$ and $K_2$ are closed. We also know, that intersections of closed subspaces are closed, so $K$ is closed in $\Omega$. We also know, that $K \subseteq K_1$. We can apply this theorem, if we can show that $K$ is closed in $K_1$. 
Question: how do we do this? Can we do this at all, given the information above?
Furthermore, I was wondering whether I'm on the right track, or if there's some other, perhaps "easier" way to show that $K$ is compact?
 A: You are on the good way, keep on ! A compact set is closed and a closed set in a compact set is compact (all this in a metric space).
A: Directly: Given an open cover $K\subseteq \bigcup_{i\in I} U_i$, we obtain an open cover $K_1=(X\setminus K_2)\cup \bigcup_{i\in I} U_i$ of $K_1$ because $K_2$ is closed. A finite subcover herof for $K_1$ (because $K_1$ is compact) is also a subcover for $K$ because $K\subseteq K_1$. From this we can drop $(X\setminus K_2)$ because it is disjoint to $K$. Hence we have a finite subcover of the original cover of $K$, i.e. $K$ is compact.
Along your path (shorter as it uses helpful lemmas): By definition of subspace topology on a subspace $A$ of $X$, a set $B\subseteq A$ is (relatively) closed iff $B=A\cap C$ for some closed subset of $X$.
A: You are right. However you should mention that $X$ is Hausdorff, since $X$ is Metrizable.

Note that not every compact subset is closed.

Example 1: Let $X= \mathbb R$ with finite complement topology. Note that $X$ is $T_1$, not $T_2$. It is not difficult to prove that $X$ is compact. Now let $S=\{0,1,2,\dots,n,\dots\}$. Note that $S$ is compact, however it is not closed. 

However it is true for a Hausdorff space.

Theorem 2: Every compact subspace of a Hausdorff space $X$ is a closed subspace of $X$.

Proof: Let $A$ be a compact subspace of $X$.  For every $x\in X\setminus A$ there exists an open set $V\subset X$ such that $x \in V$ and $A \cap V=\emptyset$, so that $X \setminus A$ is an open subset of $X$.
A: *

*$\mathrm{Closed}(A), \mathrm{Closed}(B) \implies \mathrm{Closed}(A \cap B)$

*In metric spaces: $\mathrm{Compact}(A) \implies \mathrm{Closed}(A)$

*$\mathrm{Closed}(A), \mathrm{Compact}(B), A \subset B \implies \mathrm{Compact}(A)$ 

*$A \cap B \subset A$, $A \cap B \subset B$


So


*

*$\mathrm{Compact}(K_1), \mathrm{Compact}(K_2)$

*$\mathrm{Compact}(K_1) \implies \mathrm{Closed}(K_1)$ 

*$\mathrm{Compact}(K_2) \implies \mathrm{Closed}(K_2)$

*$\mathrm{Closed}(K_1), \mathrm{Closed}(K_2) \implies \mathrm{Closed}(K_1 \cap K_2)$ 

*$\mathrm{Closed}(K_1 \cap K_2), \mathrm{Compact}(K_1), K_1 \cap K_2 \subset K_1 \implies \mathrm{Compact}(K_1 \cap K_2)$

