$K$ compact $\subseteq A_1 \cup A_2$ open, then $\exists K_1 \subseteq A_1, K_2 \subseteq A_2$ compact s.t. $K = K_1 \cup K_2$ Let $K \subseteq \mathbb R^n$ be compact, $A_1, A_2$ open sets s.t. $K \subseteq A_1 \cup A_2$. Are there $K_1 \subseteq A_1, K_2 \subseteq A_2$ compact sets s.t. $K = K_1 \cup K_2$?
 A: Yes. At each point of $K$, take an open ball a small radius centered at that point such that the ball of double radius centered at the same point entirely lies in either $A_1$ or $A_2$. Take a finite cover of $K$ by the smaller balls. The closure  of each of these smaller balls still lies in $A_1$ or $A_2$. Take $K_1=K\cap$ the union of the closures of the smaller balls lying in $A_1$ and $K_2=K\cap$ the union of the closures of the smaller balls lying in $A_2$. (The unions being finite, they are closed.) 
A: To put it more generally: let $X$ be a locally compact Hausdorff space. Let $K \subset X$ be compact and $A_1, A_2$ open such that $K \subset A_1 \cup A_2$. 
For each $x \in K$, pick $i(x) \in \{1,2\}$ and $O_x \subset X$ open such that $\overline{O_x}$ is compact and $x \in O_x \subset \overline{O_x} \subset A_{i(x)}$, which can be done by local compactness (with regularity, which is why we need Hausdorff). 
As $K$ is compact, there are finitely many $O_x$, say $O_x$ for $x \in F$, where $F \subset K$ is finite, that also cover $K$. 
Define for $i=1,2$, define $K_i = \cup \{ \overline{O_x} \cap K : x \in F, i(x) = i \}$.
Then $K_1 \cup K_2 = K$, as both unions in the definitions at least contain all $O_x$ with $x \in F$. As closed subsets of $K$, they are both compact. And $K_i \subset A_i$ as $\overline{O_x} \subset A_i(x)$ for all $x$. 
The idea is as in @Vladimir's answer, but in a more general context (and more formally written).   
