$X$ Hausdorff $\Leftrightarrow$ $\forall$ disjoint compacts $K_1,K_2:\exists$disjoint open sets $U,V$ including $K_1,K_2$ respectively Show that a topological space $X$ is separated (Hausdorff) if and only if for all compact subspaces $K_1$ and $K_2$ of $X$, there are disjoint sets $U$ and $V$ of $X$ such that $K_1 \subseteq U$ and $K_2 \subseteq V$.
I have one way ($\Leftarrow$), but I'm having some problems finding a neat proof from left to right. I think I've found a solution, but it's quite tedious. Here it is:
We assume $X$ is separated and we take two disjoint compact subspaces $K_1$ and $K_2$ of $X$.
Let's fix $x \in K_1$. Then $\forall y \in K_2$, there are two disjoint open sets $V_{x,y}$ and $U_{x,y}$ including $x$ and $y$ respectively. All these $U_{x,y}$ cover $K_2$ and therefore we can extract a finite number of them that cover it too $\{U_{x,y_i} : i = 1,...,n\}$ ($n \in \mathbb{N}^{\ast}$), and we take the correspondent $\{V_{x,y_i} : i = 1,...,n\}$. Then the sets $V_x:= \bigcap_{i = 1}^n V_{x,y_i}$ and $U_x := \bigcup_{i = 1}^n U_{x,y_i}$ are open disjoint sets.
Now I do this for all $x \in K_1$ and get a set $\{V_x : x \in K_1\}$ of open sets covering $K_1$, and corresponding $\{U_x : x \in K_1\}$ such that $\forall x \in K_1, V_x \cap U_x = \emptyset$. Since $K_1$ is compact, I now extract a finite number of those open sets $\{V_{x_i}: i=1,...,m\}$ ($m \in \mathbb{N}$), which still cover $K_1$ and the corresponding $\{U_{x_i}: i=1,...,m\}$.
Now $V :=\bigcup_{i = 1}^m V_{x_i}$ is open and covers $K_1$. While $U :=\bigcap_{i = 1}^m U_{x_i}$ is open too and covers $K_2$. Moreover $U \cap V = \emptyset$. That finishes the proof.
Is this approach correct? But more importantly, isn't there a less tedious way?
 A: It's almost correct:
You have the $U_x$ and $V_x$ such that $x \in V_x$ and $K_2 \subseteq U_x$ and $U_x \cap V_x = \emptyset$.
You then have $V_x$ for every $x \in K$ and have finitely many that cover $K_1$. You are correct in using the union of the $V_x$ to cover $K_1$, but you need the intersection (!) of the corresponding $U_x$, which being a finite intersection, is still an open neighbourhood of $K_2$. The union won't do. Check the details: suppose $x$ is in the intersection of $U$ and $V$, where $U$ is defined as the intersection, then $x$ is in some $V_{x_i}$ but the intersection garantuees it is also in the same-indexed $U_{x_i}$ as well, getting an immediate contradiction. Otherwise we'd only know $x$ is in some $V_{x_j}$ and we'd know nothing. It works the same in the firts case, that you did get correct. It's directly analogous.
This is the standard way. Often the point vs compact case is a sublemma proved separately (namely that in a Hausdorff space we can separate a point and  a compact set). I wouldn't know of any essentially simpler way. 
