If $K\subset \mathbb{R}^n$ compact and $U\subset \mathbb{R}^n$ open, $K \subset U$.Prove that $U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$ 

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.

How do you know that $U-K$ is not empty? That is where compactness of $K$ and conncetedness of $\mathbb R^{n}$ come in.
Here is a complete argument:
If $U \cap (\mathbb R^{n}-K)$ is the empty set than $U \subseteq K$. But it is given that $K \subseteq U$ so we get $U=K$. But $\mathbb R^{n}$ is connected, so the only open and closed sets are the empty set and $\mathbb R^{n}$. Since $U$ is given to be non-empty we get $K=U=\mathbb R^{n}$. But this contradicts the fact that $\mathbb R^{n}$ is not compact.
