Let $K$ be a compact subset of a Hausdorff space $X$, and $x \not \in K$. Then there are disjoint open sets $U,V$ with $K \subseteq U$ and $x \in V$.
The proof of this result that I have seen uses Choice with the Hausdorff property applied for each element of $K$. Is there a way to avoid using Choice, and if not, what form of Choice is actually necessary?
There's a choiceless motto: If you can't choose, be greedy and take everything.
Take the open cover of $K$ given by $\{U\mid U\text{ is open}, U\cap K\neq\varnothing, \exists V\text{ open s.t. }x\in V, V\cap U=\varnothing\}$. So instead of choosing an environment, take all of them.
By compactness, there is a finite subcover, $U_0,\dots,U_{n-1}$. But then $U=\bigcup_{i<n}U_i$ is also an open set in our cover. Now for each $U_i$ there is some open $V_i$ such that $x\in V_i$ and $V_i\cap U_i=\varnothing$. Set $V=\bigcap_{i<n}V_i$, then $V$ is open, $x\in V$, and $V\cap U=\varnothing$, as wanted.