Let $K\subset U\subset \mathbb R^n$, $K$ is compact and $U$ is open, bounded, connected with connected complement. I want to find an open connected set $W$ with $\mathbb R^n\setminus\operatorname{cl} W$ connected such that $K\subset W\subset \operatorname{cl} W\subset U$ and the boundary of $W$ is $(n-1)$-dimensional manifold.
I know it can be done for $n=2$, because then $U$ is homeomorphic to $\mathbb R^2$, say $h\colon U\rightarrow \mathbb R^2$ is a homeomorphism. Then $h[K]$ is contained in some open ball $B$ and $h^{-1}[B]$ is the desired set $W$.
My approach for arbitrary $n$
Cover $K$ with a finite number of balls $B_1,\ldots, B_n$ each contained in $U$ with the closure. Connect each pair $B_i$, $B_j$ with an arc $A_{ij}$ contained with $U$. Cover the union $\bigcup_{i,j}A_{ij}$ with a finite number of balls $D_1,\ldots, D_m$ each again contained in $U$ with the closure. The set $V=B_1\cup\dots\cup B_n\cup D_1\cup\dots\cup D_m$ is connected, contained in $U$ with the closure and contains $K$. Now we could fill the holes in $V$ to make the complement connected, that is, take the set $V^*=\mathbb R^n\setminus C$, where $C$ the unbounded component of $\mathbb R^n\setminus V$. However, it does not guarantee that $\mathbb R^n\setminus\operatorname{cl} V^*$ will be connected. We could also fill the holes another way, that is, take the set $V^{**}=\mathbb R^n\setminus \operatorname{cl} C'$, where $C'$ is the unbounded connected component of $\mathbb R^n\setminus \operatorname{cl} V$. But this time I'm not sure whether $V^{**}$ is connected. Note that we keep the inclusion $\operatorname{cl} V^{**}\subset U$.
It seems to me that $V^{**}$ will be connected because $V$ is a finite union of balls (and I know counterexamples with $V^{**}$ disconnected without that assumption). However, the proof I tried to produce was starting to be very involved (and still not successful).
I wonder if there is a simpler approach than mine.
Edit.
This is what I gathered from comments by Moishe Kohan.
Step 1. Cover $K$ with a finite number of closed balls $B_1,\ldots, B_n$ each contained in $U$. Connect each pair $B_i$, $B_j$ with an arc $A_{ij}$ contained with $U$. Put $K^*=B_1\cup\dots\cup B_n\cup\bigcup_{i,j}A_{ij}$. This way we enlarged $K$ to connected compact set still contained in $U$.
Step 2. Show that there exists a decreasing sequence $(U_n)$ of open connected sets with regular boundary such that $K^*\subset U_n \subset B(K^*,\frac1n)$ (I think I can fill the details using Sard's theorem). Hence $K^*=\bigcap_n\operatorname{cl}{U_n}\subset U$ and by compactness one the $U_n$'s is contained in $U$ with the closure. Set this $U_n$ to $W$.
Step 3. Replace $W$ by $W^*$ or $W^{**}$ described above, because we still need $\mathbb R^n\setminus \operatorname{cl}W$ to be connected. Intuitively, it should follow from regularity of the boundary that $W^{*}=W^{**}$ and also that the boundary of $W^*$ is still regular (maybe we will need $W$ to be reguarly open)
At the moment I can't prove the claims in step 3.
