$\epsilon$ fattening of an open set needn't be open

Question

This was a question on a test.

Let $(X,d)$ be a metric space and $\epsilon \geq 0$. Define the $\epsilon$-fattening of a set $S$ as $$S_\epsilon:=\{ x \in X; \exists s \in S: d(s,x) \leq \epsilon \}$$

As the title says, but I can't find a counterexample. It seems to me that if $S$ doesn't contain its boundary then the "fattened" set shouldn't either by the definition. Any help would be appreciated.

Answer 1

Let $X = [0,1] \cup [2,3]$ with the usual metric of $\mathbb R$. Then $S = [0,1]$ is open in $X$, but its $1$-fattening is $[0,1] \cup \{2\}$ which is not open in $X$.