How to show distance between boundary of an open set and a compact subset is positive? 
Let $n \in \mathbb N$ and $U\subset \mathbb R^n$ be an open set.Let $K\subset U$ be a compact subset of $U$. Then How can we show that distance between $K$ and $\partial U$ is positive?

Please, someone, give some hints. to solve this.
Thank you.
 A: $U$ being open $U$ and $\partial U$ are disjoint and $K\subset U$, so $K$ and $\partial U$ are disjoint too. Also $K$ and $\partial U$ are both closed. Now $d(\partial U,K)=0$ would imply $\exists~\{x_n\}\in\partial U$ and $\{y_n\}\in K$ such that $d(x_n,y_n)\to0$. As $K$ is also bounded, $\{y_n\}$ has a limit point $y\in K$. But that would imply $d(x_n,y)\to0$, which would imply $x_n\to y$, implying $y\in\partial U$, a contradiction. Hence $d(\partial U,K)>0$.
A: For $U\subset \mathbb R^n $ open, $\partial U=\bar U\cap\bar {U^c}\subset U^c $, since $U^c $ is closed.  
If $d (\partial U,K)=0$, there will be a sequence $(x_i) $ in $K $ with  $\lim_{i\to\infty }d (x_i, \partial U )=0$.  Assuming we are in a metric space,  by compactness of $K $  this sequence has a subsequence which converges to a point  $x\in K $.  This contradicts $K\subset U $.  For necessarily $x\in U^c $ ($d (x,\partial U )=0$, hence $x $ is a limit point of $U^c $).   
A: Let $\delta = \sup \{ r\ge 0 | \forall x \in K,B(x,r) \subset U \}$.
If $r =0$, Then for all $n \in \mathbb{N}$ there is some $x_n \in K$ such that $B(x_n,{1 \over n}) \not\subset U$. Since $K$ is compact,
we have $x_{n_k} \to x \in K$ for some subsequence $x_{n_k}$. However,
since $x \in K \subset U$, there is some $r'>0$ such that $B(x,r') \subset U$, which contradicts the construction of $x_n$. Hence $r>0$.
Suppose $x \in K$ and  $b \in \partial U$. Since $b \notin U$ and
$B(x, {r \over 2}) \subset U$, we see that $d(x,b) \ge {r \over 2} >0$.
