$X\neq \varnothing$ topological space that is Compact and Haussdorf without an isolated elements. I need to proof that for Every open subset $U \subset X $ and for every $x\in X$ exist open subset $V\subset X $ such that $V\subset U$ but $x\not \in \overline V$
I tried to take 2 points one x and the second y while U is some open neighberhood of y that $x\not\in U$ but because Haussdorf x have neighborhood S such that $S\bigcap U \neq \varnothing$ I stucked here...
Let $U\subseteq X$ be open, $x\in X$. Since $X$ does not have isolated points then in particular every open subset has at least two points and thus there exists $y\in U$ such that $y\neq x$. Since $X$ is Hausdorff then $x$ and $y$ can be separated by two open neighbourhoods, say $V_x, V_y$. Obviously $x\not\in \overline{V_y}$ because $\overline{V_y}\subseteq X\backslash V_x$.
Put $V:=V_y\cap U$. Then $V\subseteq U$ and $x\not\in\overline{V}\subseteq\overline{V_y}$.
Side note: I don't see how compactness plays a role here.