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Let $K\subset \mathbb{R}^n$ be compact subspace and let $\ U\subset \mathbb{R}^n\ $ be open, non-empty and containing $\ K.\ $ Prove that $\ U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$

**Attempt**

The first idea that comes to my mind was start to compare $U$ with $\left(\mathbb{R}^n -K \right)$ and see if is possible that the sets have a intersecton.

Let $U$ open notice that $\mathbb{R}^n-U \subset \mathbb{R}^n-K$ since $U$ is open
then $\mathbb{R}^n-U$ is closed. Now since $K \subset U$ then $U-K \neq \emptyset $ and
in fact $U-K \subset \mathbb{R}^n-K$.From here I affirm that $U\cap \mathbb{R}^n-K$. The problem is that I never use that $K$ is compact.
Someone can give me a hint or advice at the moment of prove future propositions that involve Complements.