Can the supports of disjoint continuous functions on a compact Haussdorf space always be separated by open sets?
I.e.: given a compact Haussdorf space $X$ and a sequence of continuous functions $h_k$ on $X$, do disjoint open sets $O_k$ exist such that $\operatorname{supp}h_k \subseteq O_k$?
Yes, assuming the codomain of these functions has nice properties and that you only have a finite collection of functions. Usually, if you are going to talk about support, your codomain is a vector space. If this is the case, or even if your codomain is just Hausdorff, and you generalize the meaning of support, the answer is yes.
If $f$ and $g$ are two continuous functions $X\to Y$ where $X$ is a compact Hausdorff space, and $Y$ is a Hausdorff space such that there is a point $y\in Y$ such that $\overline A=\overline{f^{-1}(Y\setminus\{y\})}$ and $\overline B=\overline{g^{-1}(Y\setminus\{y\})}$ are disjoint. If $Y$ is a vector space, and $y=0_Y$, then $\overline A$ and $\overline B$ are the supports of $f$ and $g$. Since Compact Hausdorff spaces are normal, there exist disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively.
This proof can be extended for any finite collection of functions quite easily. But if you have an infinte sequence of functions, the answer is no. Consider $h_1:\mathbb R^2\to\mathbb R$ whose support is exactly $\{(x,y):x\leq 0\}$, and then $h_n$ a bump function whose support is the closed disk of radius $1/2^{n+2}$ centered at $(1/2^n,0)$ for all $n>1$. Then the supports are all disjoint, but every open neighborhood of the support of $h_1$ contains all but finitely many supports of the other $h_n$.