separating disjoint compact set and closed set in regular space 
In a regular space $X$, a compact set $A$ and a disjoint closed set $F$ can be separated by disjoint open sets $U,V$.

Since $X$ is regular, for every $x\in A$ there exists disjoint open sets $U_x,V_x\subset X$ such that
$$x\in U_x,\ x\in V_x,\text{ and } U_x\cap V_x = \emptyset.$$
Clearly, $A\subset\bigcup_{x\in A}U_x$, so there exists a finite set $\{x_1,x_2,...,x_k\}\subset A$ such that $A\subset\bigcup_{i=1}^{k}U_{x_i}$. Let $U:=\bigcup_{i=1}^{k}U_{x_i}$ and $V:=\bigcap_{i=1}^{k}V_{x_i}$. Clearly $A\subset U$ and $U\cap V=\emptyset$ but I'm not sure why $F\subset V$.
 A: Apply regularity to $F$ and all $a \in A$:
We have open $U_a,V_a$, such that $a \in U_a, F \subseteq V_a$ and $U_a \cap V_a= \emptyset$. This we do for each $a \in A$.
The $\{U_a\mid a \in A\}$ cover the compact $A$ so finitely many do too:
$$\exists a_1, \ldots,a_n \in A: A \subseteq V:= \bigcup_{i=1}^n U_{a_i}$$
Then define $$U= \bigcap_{i=1}^n V_{a_i}$$
and as for all $a$, $F \subseteq V_a$ we have $F \subseteq U$ and $U$ is open because we have a finite intersection of opens.
Then $U \cap V = \emptyset$: suppose not, then we have some $x \in U \cap V$ and we have that $x \in V$ so for some $a_i$, $x \in U_{a_i}$. But also $x \in V \subseteq V_{a_i}$ and this contradicts the disjointness of all corresponding $U_a$ and $V_a$. So No such $x$ exists and the sets are disjoint.
Your proof was that in a Hausdorff space we can separate a point $x$ from a disjoint compact set, and you needed to intersect the $U_{x_i}$ and union the $V_{x_i}$, not the other way round. Disjointness is proved the same way as above.
