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.
 A: 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.
A: Let $\ U\ne\emptyset\ $ be open. If $\ U\ $ were also compact then it would be closed (and open at the same time). Then $\ U\ $ must be empty (but it is not!) or $\ U=\mathbb R^n\ $ because $\ \mathbb R^n\ $ is connected. But $\ \mathbb R^n\ $ is not compact -- a contradiction.
It follows that non-empty open $\ U\ $ is not compact. If $\ K\subseteq U\ $ for a compact K then (since $\ K\ne U)$ we obtain $\ U\setminus K\ne\emptyset.$
A: Hint
Supose $U \cap \left(\mathbb{R}^n-K\right)=\emptyset$
A: Method 1. $\Bbb R^n$ is a connected space. It is also a non-compact Hausdorff space, so a compact  $K$ is   closed, and  $\Bbb R^n\setminus K$ is non-empty & open.
For any $v\in \Bbb R^n$ we have $v\not\in U\implies v\not \in K\implies v\in \Bbb R^n\setminus K.$ So $$\Bbb R^n=U\cup ( \Bbb R^n\setminus K).$$ But now if  $U\cap (\Bbb R^n\setminus K)$ is empty then $\Bbb R^n$ is the union of the disjoint pair $\{U,\, \Bbb R^n\setminus K\}$ of non-empty open sets, contrary to the connectedness of  $\Bbb R^n.$
This will hold verbatim under the weaker condition on $K$ that $\overline K=K\ne \Bbb R^n.$
Appendix. The connectedness of $\Bbb R$ implies that a path-connected space is a connected space. Now if $u,v\in \Bbb R^n$ then $f(r)=(1-r)u+rv$ is continuous from $[0,1]$ into $\Bbb R^n$ with $f(0)=u$ and $f(1)=v.$ So  $\Bbb R^n$ is path-connected.
Method 2. (i). If $U=\Bbb R^n$ then $U\cap (\Bbb R^n\setminus K)=\Bbb R^n\setminus K\ne \emptyset.$
(ii). If $\emptyset \ne U\ne\Bbb R^n$ then $\overline U\ne U$ (...because $U$ is open and $\Bbb R^n$ is connected...) so let $p\in \overline U\setminus U.$ Then $p\not \in \overline K$ (...because $p\not\in U$ and $\overline K=K\subset U$...) so there exists an open $V$ with $p\in V$ and $K\cap V=\emptyset.$ But $p\in \overline U$ so there must exist $q\in V\cap U.$ So $q\in U\cap (\Bbb R^n\setminus K).$
A: How come we don't have the simplest answer?
Both $\ U\ $ and $\ \mathbb R^n\setminus K\ $ are open, and non-empty and
$$ U\cup(R^n\setminus K)\ =\ R^n $$
But $\ \mathbb R^n\ $ is connected hence
$\ U\cap(R^n\setminus K)\ \ne\emptyset.$   Done.
