Let $\Omega$ be an open subset of $\mathbb{R}^n$, and let $A\subset\Omega$ be a non-empty connected open set. Assume that there exists a constant $R>0$ such that $$ (1) \qquad \text{dist}(x, \partial A) = \text{dist}(x, \partial \Omega) - R, \qquad \forall x\in A, $$ where $\text{dist}(x, K) := \min_{y\in K} |x-y|$ for every closed set $K\subset \mathbb{R}^n$.
I would like to prove (or disprove...) that $A$ is a connected component of the set $\Omega_R := \{x\in\Omega:\ \text{dist}(x, \partial\Omega) > R\}$.
It is not difficult to show that $A\subseteq\Omega_R$.
In particular, if $B$ is a connected component of $\Omega_R$ such that $A\cap B\neq\emptyset$, then $A\subset B$, and we have to prove that $A=B$.
Since now, I have proved the following property that seems to be useful: $$ (2)\qquad x\in A,\ y\in \Pi_{\partial A}(x) \quad\Longrightarrow\quad z := y + R\, \frac{y-x}{|y-x|}\in\Pi_{\partial\Omega}(x), $$ where $\Pi_K(x)$ is the set of projections of $x$ on $K$.
Any hint will be appreciated.
Ok, try following: for every boundary point $y \in \partial A$ take a look at a point $x\in A$ which is really close to it, say $\text{dist}(x,y) = \varepsilon$. You will get that distance from $x$ to $\partial \Omega$ is not more than $R+\varepsilon$ and hence the distance from $y$ to $\partial \Omega$ in not more than $R + 2\varepsilon$.
Now consider $\varepsilon \to 0$, you will get that $\partial A \subset \{x\in \Omega \mid \text{dist}(x,\partial \Omega)\le R\}$. Therefore $\partial A \cap \{x\in \Omega \mid \text{dist}(x,\partial \Omega)> R\} = \partial A \cap \Omega_R = \emptyset$.
From here you can probably prove what you need.