Showing a space is not Hausdorff A space is Hausdorff if given two different points $x$ and $y$, there are two open disjoint sets $G_x$ and $G_y$ such that $x\in G_x$ and $y\in G_y$. Prove that if $K$ is compact and $K$ is not closed, then the universe can not be Hausdorff.
I am having some issues trying to start this proof because I think it should be done by contraposition but I do not know what "can be Hausdorff" means in a proof. Otherwise, doing it directly, I do not know exactly how to use the idea of not closed. My thoughts were to somehow use the complement and say $ K^c $ is not open. Ideas? Thank You!
 A: What you can show is a bit stronger than "every compact subset of Hausdorff space is closed". You can show that in Hausdorff space, points can be separated from compact sets by disjoint open sets.
Fix a point $x\not\in K$ and for every $y\in K$ choose disjoint open $U_y\ni x$  and $V_y\ni y$. $\{V_y\}$ is obviously an open cover for $K$, so there is finite subcover $\{V_{y_1},\ldots,V_{y_n}\}$. Define $U = U_{y_1}\cap\ldots\cap U_{y_n}$ and $V = V_{y_1}\cup\ldots\cup V_{y_n}$. Then $x\in U$, $K\subseteq V$ and $U\cap V=\emptyset$.
In particular, $K^c$ is open.
Side note. If you use the above result, with the same reasoning you can show that any two disjoint compact sets can be separated by disjoint open sets.
A: The usual way to do this is to show that in a Hausdorff space a compact set is closed.
A: Suppose universe space is hausdorff. Take $y \in K^c$ and by hausdorffness for each $x \in K$ there are neighbourhoods $U_x$ and $V_x$ of $x$ and $y$ respectively which are disjoint. So, these open set $\{U_x\}_{x \in K}$ cover $K$. Since $K$ is compact it has a finite subcover $\{U_{x_1}, U_{x_2},...,U_{x_n}\}$. Now take $V = V_{x_1}\cap V_{x_2}\cap ... \cap V_{x_n}$ which is a non empty neighborhood of $y \in K^c$ and also $V$ is disjoint from $K$ as it disjoint from each $U_{x_i}$. So, $K^c$ is open which is a contradiction.
