Does exist a compact connected $K' \subset U$ such that $K \subset K'$, if $U$ is an open connected and $K$ in $U$ a compact? Let $U$ be an open and connected set in $\mathbb{R}^n$.
Suppose $K \subset U$ is a compact set. Is it true that there exists a compact and connected set $K' \subset U$ such that $K \subset K'$?
I know that there exists a compact set $K' \subset U$ such that $K \subset K'$.
But how I can guarantee that $K'$ is connected? 
 A: Since $K$ is compact, there exists $\epsilon > 0$ such that $\bigcup_{x \in K} B(x,\epsilon) \subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},\ldots,x_{n} \in K$ such that $K \subset \bigcup_{i} B(x_{i},\epsilon/2) \subset \bigcup_{i} \overline{B(x_{i},\epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $\gamma_{i,j} : [0,1] \rightarrow U$ such that $\gamma(0) = x_{i}$ and $\gamma(1) = x_{j}$.
Then
$K' = \bigcup_{i} \left( \overline{B(x_{i},\epsilon/2)} \cup \bigcup_{j > i} \gamma_{i,j}([0,1]) \right) \subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
A: For any $x\in K$ there is $\epsilon >0$ such that $B(x,\epsilon)\subset U$. Let $A_x=B(x,\epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$. 
Since $\{A_x, x\in K\}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,\dots,A_n$. Choose $x_0\in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $\mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $\overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets. 
By construction $K'\subset U$.
