Complement of small enough $\varepsilon$-neighbourhood of boundary is connected 
Let $U \subset \mathbb{R}^n$ be an open, bounded and connected subset of $\mathbb{R}^n$ for which $\partial U$ is smooth. I would like to show that for $\varepsilon > 0$ small enough, the set $$U_\varepsilon := \{x \in U \ \mid \ d(x, \partial U) > \varepsilon \} $$ is connected. However, I have not managed to do so.

I have seen that there are counter-examples (Is for open connected $U$ the set $U_\varepsilon$ for small $\varepsilon$ connected?) if the assumption that $\partial U$ is smooth is dropped. However, I do not see how the assumption that $\partial U$ is smooth can be used to prove that $U_\varepsilon$ is connected. I suppose that this is the same as proving that the complement of a collar neighbourhood of $\partial U$ is connected (hence why $\partial U$ needs to be smooth), but I do not know how to proceed.
This question comes up when proving, for instance, that if $u$ is a Sobolev function on $U$ with vanishing weak gradient, then it must be constant almost everywhere (the argument involves mollifying $u$ with a cut-off function, and the resulting function has vanishing weak gradient and is defined only on $U_\varepsilon$).
 A: Yes, that holds true.
Since $U$ is open and connected it is also path connected.
Let $N$ be a unit normal vector field on $\partial U$, pointing inside $U$ (for simplicity assume that this direction is unique). Consider for $\delta>0$ the map
$$ \phi : T=T_\delta=(q,t) \in \partial U \times (0,\delta] \to 
   q+tN(q) \in {\Bbb R}^n. $$
Using that $\partial U$ is smooth and compact, we may find $\delta>0$ small enough so that the map $\phi$ has the following properties:

*

*For every $(q,t)\in T$: $d(\phi(q,t),\partial U)=t$.

*$\phi$ is a diffeo from $T$ onto $\phi(T) = U\setminus U_{\delta}$.

(sketch of proof below)
Define for $0<r<\delta$ the projection map: $\psi_r: U \mapsto \overline{U_r}$ given by $\psi_r(x)=x$ for $x\in \overline{U_r}$ and when $x=\phi(q,t)$ with $(q,t)\in T$ and $0<t<r$ we set $\psi_r(x)=\phi(q,r)$.
The map $\psi_r$ is Lipschitz continuous.
Claim 1: $\overline{U_r}$ is path connected. Proof: Let
$x,y\in \overline{U_r}$ and let $\gamma:[0,1]\to U$ be a path connecting $x$ and $y$ in $U$. Then $\psi_r\circ \gamma$
is a path connecting $x$ and $y$ in $\overline{U_r}$.
Claim 2:  $U_\epsilon$ is path connected for any $0<\epsilon<\delta$. Proof: Let $x,y\in U_\epsilon$. Then
$x,y\in \overline{U_r}$ for some $r\in(\epsilon,\delta)$. Then the path constructed in Claim 1 will also connect $x$ and $y$ in $U_\epsilon$.

Proving points 1. and 2. concerning $\phi$ are quite messy (at least I don't know any short and easily readable proofs).
Some steps in a proof could look like:
-- Local coordinates:  Pick $q_0\in\partial U$ and an o.n.b for $T_{q_0}\partial U$:
$e_1,...,e_{n-1}$.  Together with $e_n=N(q_0)$ we obtain an o.n.b for ${\Bbb R}^n$.
-- In this basis we write $(x,t)=(x_1,...,x_{n-1},t)\in {\Bbb R}^n$ for
the coordinates. In a small enough neighborhood of $q_0$, the boundary is
the graph of a map $t=f(x_1,...,x_{n-1})$ for which $f(0)=0$ and $\nabla f(0)=0$. We assume $f \in C^2$.
A point on the
surface is given by $q(x)=\sum_{k=1}^{n-1} x_k e_k+f(x)e_n$ and finally in these local coordinates the map $phi$ takes the form:
$\phi(q(x),t)= q(x) + t N(q(x))$.
You verify that $\frac{\partial \phi}{\partial(x,t)}_{(0,0)} = (e_1,...,e_n)$ which is invertible. Thus $\phi$ defines a local diffeomorphism.
Now, using that $f$ is $C^2$ we may show that for $t>0$ small enough,
$B(q+tN(q),t)\subset U$ and
$\partial B(q+tN(q),t)\cap \partial U =\{q\}$.
($t$ should be smaller than the smallest radius of curvature
of $\partial U$ in a neighborhood of $q$,
but you also have to worry about the distance to other
parts of $\partial U$).
Then $q\in \partial U$ is the unique point in the
boundary closest to $q+tN(q)$ and their distance is precisely $t$.
Using compactness (and  choosing $\delta>0$ small enough), $\phi$ becomes a well defined map on $T_\delta$ and verifies the stated
properties.
