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? 

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 $K'\subset U$.
