Let $X$ denote a complete metric space and consider a subset $A \subseteq X$. Call $A$ uniformly isolated iff there exists $r > 0$ such that for all $a \in A$, we have that $B_r(a) \cap A = \{a\}$.
Question. If $A$ is uniformly isolated, is it necessarily closed?
I am especially interested in the case where $X$ is a Banach space.
A uniformly isolated set is closed: Let $(x_n)$ be a sequence in $A$ and suppose $x_n \to x \in X$. Choose $N \in \mathbb N$ such that $d(x_n, x) < \frac{r}3$ for $n \ge N$, hence $d(x_n, x_m) < \frac{2r}3$ for $n \ge m$. As the points of $A$ are at least $r$ apart, $(x_n)$ must be constant for $n \ge N$, so there is an $a \in A$ such that $x_n = a$ for $n \ge N$. As $x_n \to x$, we must have $x = a \in A$.
So $A$ is closed.