In this question "compact" does not include Hausdorff.
Let $X$ be a compact space and $U_1,\ldots, U_n$ be an open cover of $X$. Does there exist a compact cover $K_1,\ldots,K_n$ of $X$ such that $K_i \subset U_i$?
I doubt that it is true for arbitrary compact $X$. But obviously for Hausdorff $X$ the answer is "yes". More generally, it is true if there exists a continuous surjection $f : X' \to X$ with a compact Hausdorff $X'$.
I tried $X = \alpha(\mathbb Q)$ = Alexandroff compactification of $\mathbb Q$ which is not the continuous image of a compact Hausdorff space (see An example of a compact topological space which is not the continuous image of a compact Hausdorff space?), but not even in this case I was able to find the answer.
Suppose that $X$ is compact and $F$ and $G$ are disjoint closed subsets of $X$. Then $F^\complement$ and $G^\complement$ form an open cover of $X$. Suppose we have compact $K_1 \subseteq F^\complement$ and $K_2 \subseteq G^\complement$ such that $K_1 \cup K_2=X$, then $F \subseteq K_1^\complement$, and $G \subseteq K_2^\complement$ so that $F$ and $G$ have disjoint co-compact supersets. So clearly in a Hausdorff space (or a KC-space (where compact sets are closed, so co-compact sets are open)), the property is just stating the normality of $X$ (which is nothing new for Hausdorff $X$).
As $X=\alpha(\Bbb Q)$ is a compact, KC space, it cannot have the cover-property as it's not normal.