Compact Subset of $\mathbb{R}^n$ Let $U$ be a proper open subset of $\mathbb{R}^n$ and $K$ be a compact subset of $U$. Show that there exists $r$ $>$ $0$ such that the set $\{ y$ $\in$ $\mathbb{R}^n : ||y-x|| \leq$ $r$ for some $x$ $\in$ $K$} is a compact subset of $U$.
My idea: $d(K,U^{c})$ $=$ $r$ $>$ $0$. I think this is the $"r"$ that we are actually looking for. But I cannot complete the proof rigorously. Thanks for any help.
 A: Your idea is good, but the number you're considering is too large.
Part 1 (what you probably already have)
Since $K$ and $U^c$ are disjoint closed sets and $K$ is compact, the set
$$
\{d(x,y):x\in K, y\in U^c\}
$$
has a positive minimum. Indeed, for each $x\in K$, 
$$
d(x,U^c)=\inf\{d(x,y):y\in U^c\}
$$
is actually a minimum, because $U^c$ is closed. The function $x\in K\mapsto d(x,U^c)$ is continuous (easy proof), so it has a minimum, as $K$ is compact. Call this minimum $2r$. We have $r>0$: indeed, let $x_0\in K$ with $d(x_0,U^c)=2r$. Since $x_0\in U$, there exists a ball $B(x_0,\delta)\subseteq U$, so $B(x_0,\delta)\cap U^c=\emptyset$ and $d(x_0,U^c)\ge\delta$.
Part 2 (the construction)
Now suppose $y\in\mathbb{R}^n$ and $d(x,y)\le r$, for some $x\in K$. If $y\in U^c$, then, by definition, $d(x,y)\ge 2r$: a contradiction. [This shows why your conjecture was off.]
Therefore
$$
K'=\{y\in\mathbb{R}^n:d(x,y)\le r\text{ for some }x\in K\}\subseteq U
$$
The set $K'$ is bounded. Indeed, if $d(x,x')\le k$, for every $x,x'\in K$, then
$$
d(y,y')\le k+2r
$$
for every $y,y'\in K'$, by the triangle inequality.

 Let $d(x,y)\le r$ and $d(x',y')\le r$, with $x,x'\in K$; then$$d(y,y')\le d(y,x)+d(x,x')+d(x',y')\le r+k+r$$

Finally, $K'$ is closed.

 Let $(y_n)$ be a sequence in $K'$ convergent to $y$. We need to prove that $y\in K'$. For every $n$, fix $x_n\in K$ with $d(x_n,y_n)\le r$. Since $K$ is compact, there is a subsequence $(x_{n_k})$ convergent to $x\in K$. Why is $d(x,y)\le r$?

