If $C_\varepsilon(p)$ is connected, is $B_\varepsilon(p)$ connected? Let $(X,d)$ be a metric space, $p\in X$ and $\varepsilon>0$. Suppose that the closed ball $C_\varepsilon(p)$ is connected. Must the open ball $B_\varepsilon(p)$ be connected?
I'm not sure if this statement is true, but I haven't found any counterexamples. So, I've been trying to prove it. I've tried to prove that if $B_\varepsilon(p)$ has a non-trivial clopen subset, then $C_\varepsilon(p)$ has to have a non-trivial clopen subset, but I haven't had any success.
I would really appreciate it if someone either helps me prove this statement or find a counterexample.
 A: Take the set of all points in $\mathbb{R}^2$ where at least one coordinate is an integer. This gives you a grid. Define the distance between two points as the shortest path length between the two points. This is a metric space, let's call it $X$.
Now remove the point $(0,1)$ from this space. The remaining space with the same metric is still a metric space $X_1$.
Now the closed ball of radius $2$ around the origin in $X_1$ is connected. It is a 2 x 2 grid with one point missing. But the open ball of radius $2$ around the origin in $X_1$ is no longer connected because you now miss the $4$ corners of the 2 x 2 grid as well a the point $(0,1)$.
So for general metric spaces your claim is not true. Not that my example is Hausdorff but it is not geodesic (meaning there is a distance realizing path for any two points). The space is not compact but everything interesting happens inside a compact region, so you could easily build a compact example. I suspect that your claim is true for geodesic metric spaces but I don't have a proof for that.
