# Closed balls vs closure of open balls

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.

Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R_x|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$.
Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R_x$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$.
Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.