Show that there is a compact neighbourhood $B$ of $x$ such that $B \cap F = \emptyset$. Let $X$ be a compact Hausdorff space, $F \subset X$ closed and $x \notin F$ . Show that there is a compact neighbourhood $B$ of $x$ such that $B \cap F = \emptyset$. 
I'm trying to use the fact that compact subspaces of Hausdorff spaces are closed
this is not homework, it is from a book with no solutions
 A: Hint: Compact Hausdorff space is completely regular, so you can separate $x$ from $F$ by disjoint open neighborhoods, $U,V$ such that $x\in U$ and $F\subseteq V$. Show that the closure of $U$ is compact and disjoint of $F$.
A: $F$ is compact, because it is closed in the compact space $X$ (so we are using the general fact that "closed in compact is compact"). 
For every $y \in F$, we pick disjoint open sets $U_y$ (contains $y$) and $V_y$ (contains $x$) by Hausdorffness (here we use $x \notin F$, so that all these pairs consist of 2 points!).
Finitely many $U_y$, say $U_{y_1},\ldots,U_{y_k}$ cover $F$, by compactness. 
Then we can take $V = \overline{V_{y_1} \cap \ldots \cap V_{y_k} }$ which is a neighbourhood of $y$ (as it contains a finite intersection of neighbourhoods). And for every point $p$ in $F$ we know it is in some $U_{y_i}$, which is disjoint from $V_{y_i}$. So $V_{y_1} \cap\ldots\cap V_{y_k} \subset V_{y_i} \subset X \setminus U_{y_i}$ and as the last set is closed, $V \subset X\setminus U_{y_i}$ which means that $p \notin V$. As $p$ is arbitrary in $F$, $F$ is disjoint from $V$, as required.
