We work in a separable metric space $(X,d)$. With $\overline{B}(x,r)$ I denote the closed ball around $x$ of radius $r$, and with $cl \ B(x,r)$ I denote the closure of the open ball. Clearly, we always have $cl \ B(x,r) \subseteq \overline{B}(x,r)$, and it is easy to construct examples where the inclusion is strict.

**Question:** Can we have this locally everywhere, meaning does there exists a separable metric space $(X,d)$, a point $x \in X$ and a real $R$ such that for all $r$ with $R > r > 0$ we find that $cl \ B(x,r) \subsetneq \overline{B}(x,r)$ ?

My naive attempt to construct this by brute-force violates the separability criterion. I am tempted to believe that we can have the difference only for countably many radii, but so far I was not able to prove this.

Restrictions on $(X,d)$ I'd be willing to accept (and that I would particularly enjoy in a positive example) are completeness and local compactness.

This question here is related, it is asking about spaces where $cl \ B(x,r) = \overline{B}(x,r)$ holds always.