Compact subspace of Hausdorff space I have a problem that states:
If $X$ is Hausdorff then for every $x \in X$ and every compact subset $A \subset X$ there exists disjoint neighborhoods $U \ni x, V \supset A$.
So my question is how is this possible if $x \in A \subset X$, So am I assuming this $x \in X \setminus A$? In which case this is open since $A \subset X$ is closed. And by Hausdorfness we can find disjoint neighborhoods then intersect one and union the other to get my disjoint open balls? Am I on the right track?
 A: 
So my question is how is this possible if $x \in A \subset X$

Well, it is not possible in such case. And so there's a missing assumption: $x\not\in A$.

And by Hausdorfness we can find disjoint neighborhoods then intersect one and union the other to get my disjoint open balls?  Am I on the right track?

Almost there. You need one more step: remember that an (infinite) intersection of open subsets does not have to be open. You need to reduce this situation to finite case. And you can do that, since $A$ is compact: you can take a finite subfamily that covers whole $A$, and intersect corresponding neighbourhoods of $x$.
A: Let $A\subset X$ be compact $x\in X\setminus A$ . (If $A$ is empty then it trivially follows as $\phi$ and $X$ are such $U$ and $V$)
Pick an arbitrary $y\in A$ .
By Hausdorffness there exists $V_{y}$ and $U_{y}$ open and disjoint such that $x\in U_{y}$ and $y\in V_{y}$.
Cover $A$ up by these $V_{y}$'s . That is $\{V_{y}\}_{y\in A}$ covers $A$.
By compactness there exists $V_{y_1},...,V_{y_n}$ such that this is a finite subcover of $A$ . Correspondingly you have $U_{y_1},...,U_{y_n}$ such that $x\in U_{y_i}$ .
Then $V=\displaystyle\bigcup_{i=1}^{n}V_{y_{i}}$ and $U=\displaystyle\bigcap_{i=1}^{n}U_{y_{i}}$ are disjoint open sets such that $A\subset V$ and $x\in U$.
