Convex hull of open sets is an open set? 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.
 A: I think I have a counterexample.
Look at $X=\mathbb{R}^2\setminus\{(x,y)\in\mathbb{R}^2:|x|<1,y\neq 0\}$. Below is a picture of $X$ in red as a subset of $\mathbb{R}^2$ in A.

The metric on $X$ comes from the Euclidean metric. $X$ consists of three parts: "the left part", "the right part" and "the middle part". In each part separately, distances and metric segments are defined as induced by the intrinsic metric from $\mathbb{R}$ - see B for some examples. For two points that lie in different parts, one has to pass through one (or two) "choke points" at $Q_1, Q_2$, but except for that the metric segments consists of two or three usual Euclidean segments.
Now take as open set $S$ the union of two open balls: $S=B((-3,0);1)\cup B((3,0);1)$ for $P_1=(-3,0),P_2=(3,0)$. The convex hull $C=\text{conv}(S)$ should be $S$ together with two "open sectorlike shapes" and the "middle part" as depicted in C. If we look at the two "choke points" $Q_1, Q_2$ we see $Q_1,Q_2\in C$, however no open subset containing any of the two choke points is contained in $C$. An open ball around $Q_1$ is depicted in green in D - note that only the green part of the black disc is part of the ball.
Now this is not a formal proof, but I hope it fits. If my intuition derailed me somewhere, please object and let me know.
Remark: In this metric on $X$ all open balls should be identical to their convex hull. If we take the metric induced by the $1$-norm we get open balls that are squares oriented as "diamonds", while their convex hull should be squares oriented as "boxes". If we replace the Euclidean metric by the $1$-metric in the construction with $X$, I think that we do not get a counterexample. If the opening angle at choke points would not be $180^{\circ}$ but larger (suitably adjusting $X$), I guess one would also get a counterexample coming from the $1$-metric. By "$180^{\circ}$" I mean the usual Euclidean angles on $\mathbb{R}^2$ - nothing which has to do with the counterexample metrics.
