Boundary of an open set in a Banach space

[Principe2]

Prove (or disprove) the following statement: the boundary of an open set in a Banach space does not contain open sets.

[Principe2]

I think you are right, but I was thinking to another thing, while writing that one. So, let us state my original problem.

Let $(X,d)$ be a metric space. Let $A$ be an open and bounded subset of $X$, $r>0$, define $B(A,r)=\{x\in X : d(x,A)
1)Prove that if $X$ is a Banach space (probably normed is enough), then $N(A,r)\setminus B(A,r)$ cannot contain open sets.

2) Find an example of a metric space $X$ and a subset $A$ such that it does.

3) Find an example of metric space $X$ and a subset $A$ such that $N(A,r)\setminus B(A,r)$ contains an homeomorphic copy of $A$.

[dissonance]

I believe this claim to be true in an arbitrary topological space. Indeed, let \(U\) be an open subset of a topological space \(X\). We claim that the interior part of the boundary of \(U\) is empty, that is \(\big( \partial U \big)^\circ= \varnothing\). To see this write

\[X=U \cup \partial U \cup (U^C)^\circ \]

where \(U^C\) stands for the complement of \(U\). Suppose there exists \(x \in \big(\partial U\big)^\circ\). Since \(U\) and \(\partial U\) are disjoint, \(x \in (U^C)^\circ\). But \((U^C)^\circ\) is open and so \((U^C)^\circ\) and \(\partial U\) need be disjoint as well. We just got a contradiction.

What do you think?