Compact sets closed in Hausdorff spaces without choice? An elementary proof that compact sets are closed in Hausdorff spaces involves making arbitrary choices based on the Hausdorff property. Is there a way to avoid invoking choice?

Yes, there is.
Let $\langle X,\tau\rangle$ be a Hausdorff space, and let $K$ be a compact subset of $X$. Suppose that $K$ is not closed. Then we can pick $p\in(\operatorname{cl}K)\setminus K$. (Note that this does not require $\mathsf{AC}$: it’s a single choice.) Let 
$$\mathscr{U}=\{U\in\tau:p\notin\operatorname{cl}U\}\;.$$


*

*$K\subseteq\bigcup\mathscr{U}$.


If not, pick $x\in K\setminus\bigcup\mathscr{U}$. $X$ is Hausdorff, so there are disjoint $U,V\in\tau$ such that $x\in U$ and $p\in V$. But then $U\in\mathscr{U}$, contradicting the choice of $x$. Note that I picked only three things here, $x$, $U$, and $V$; this does not require $\mathsf{AC}$.
$K$ is compact, so there is a finite $\{U_1,\ldots,U_n\}\subseteq\mathscr{U}$ such that $K\subseteq\bigcup_{k=1}^nU_k$. Then 
$$p\notin\bigcup_{k=1}^n\operatorname{cl}U_k=\operatorname{cl}\bigcup_{k=1}^nU_k\supseteq\operatorname{cl}K\;,$$
contradicting the choice of $p$. Therefore $K$ is closed.
