Apparently, there are no such nice properties.

Even $K=[0,1]$ will not be such a compact Hausdorff space. Indeed, consider the countable double-indexed family $(U_{n,k}\colon n\in\Bbb N,k\in\{0,\dots,n-1\})$ of nonempty open subsets of $K$, where $U_{n,k}:=(\frac kn,\frac{k+1}n)$. Then any nonempty open subset $V_{1,0}$ of $U_{1,0}$ will contain entirely (infinitely many) other $U_{n,k}$'s. So, no nonempty subsets of those other $U_{n,k}$'s will be disjoint from $V_{1,0}$.

Apparently, there are no such nice properties.

Even $K=[0,1]$ will not be such a compact Hausdorff space. Indeed, consider the countable double-indexed family $(U_{n,k}\colon n\in\Bbb N,k\in\{0,\dots,n-1\})$ of nonempty open subsets of $K$, where $U_{n,k}:=(\frac kn,\frac{k+1}n)$. Then any nonempty open subset $V_{1,0}$ of $U_{1,0}=(0,1)$ will contain entirely (infinitely many) other $U_{n,k}$'s. So, no nonempty subsets of those other $U_{n,k}$'s will be disjoint from $V_{1,0}$.