Let $K \subset \mathbb R^n$ be a compact set with an open finite covering $(O_i)_{i=1,...,n}$. Then there exist compact sets $K_i \subset O_i$, $\forall i =1,...,n$ such that $K \subset \bigcup^n_{i=1} K_i$.
I came across this line of reasoning in a proof of partition of unity. However, I could not find the proof. In fact, I could prove this for open balls, not open sets.
Any hint would be highly appreciated :D. Thank you!
The proof linked in the comment requires the fact that compact $K \subset \mathbb R^n$ is normal. Here is a direct proof in $\mathbb R^n$.
We can assume that $O_i$'s are bounded (if not, take the intersection with the open set $\{ x \in \mathbb R^n: d(x,K) < r \}$, for some $r>0$) and denote their boundaries as $\partial O_i$'s. Let
$$ U^\epsilon_i := O_i \cap \{x \in R^n: d(x,\partial O_i) > \epsilon \}, \quad i =1,...,n. $$
$U^\epsilon_i$'s are open as each of them is an intersection of two open sets. Clearly for any $x\in K$, there exists $i \in \{ 1,...,n\}$ and $\epsilon>0$ such that $x \in U^\epsilon_i$. Therefore, the collection $\{U_i^\epsilon, i = 1,...,n, \epsilon >0 \}$ forms an open covering of $K$.
It will suffice if we can show
$$ \bar U^\epsilon_i := O_i \cap \{x \in R^n: d(x,\partial O_i) \geq \epsilon \}, \quad i =1,...n $$
is closed, hence compact (as $O_i$'s are bounded), because $U^\epsilon_i \subset \bar U^\epsilon_i \subset O_i$. This is not too hard, let $(x_n) \subset \bar U^\epsilon_i$ converging to $x$, by the continuity of $y \mapsto d(y, \partial O_i)$, it holds $d(x,\partial O_i) \geq \epsilon$ and clearly $x \in \bar O_i$ so $x\in O_i$, thus $x \in U^\epsilon_i$.
Finally, do some basic grouping job just like in the link at the end. The grouping of finite cover $(\bar U^\epsilon_i)$ will satisfy your need.