Let $(X,d)$ be a metric space. For all points $x,y \in X$ we define the metric segment between them as the following set:
$$\left [ x,y \right ] = \left \{ z \in X : d(x,z)+d(z,y)=d(x,y)\right \}$$
We then say that a set $S\subseteq X$ is convex if for all $x,y \in S$ it holds true that $\left [ x,y \right ] \subseteq S$.
It can be easily shown that arbitrary intersection of convex sets in metric spaces is a convex set. Therefore, for each subset $S \subseteq X$ of a metric space $(X,d)$ we define its convex hull as the set $\mathrm{conv}(S)=\bigcap_{}^{} \left \{ U \supseteq S : U \; \mathrm{convex} \right \}$.
My question is for a metric space $(X,d)$ and a open subset $S \subseteq X$, is the set $\mathrm{conv}(S)$ open?
Note: This question has been associated to my post, but it uses an inequivalent notion of convexity that cannot be used in metric spaces.