Topology basis consisting of convex sets in metric spaces 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 $\left [ x,y \right ] \subseteq S$. Denote by $\tau$ the topology on $X$ induced by the metric $d$.
My question is does there exist a family $\mathcal{B}\subseteq \tau$ of convex sets such that $\mathcal{B}$ is a basis for the topology $\tau$?
It should be noted that open and closed balls in metric spaces are not necessarily convex sets. Also, arbitrary intersection of convex sets in metric spaces is a convex set. Thus, we can define convex hulls.
 A: Let $X=\ell^1$.  I claim that the convex hull (in your sense) of any ball in $X$ is unbounded, so there are no bounded open convex (in your sense) sets in $X$.  By symmetry it suffices to consider the closed unit ball $B$ centered at $0$.  Let $e_n$ denote the sequence whose $n$th entry is $1$ and all other entries are $0$.  Then each $e_n$ is in $B$.  But if $n\neq m$, then $e_n+e_m\in[e_n,e_m]$, so $e_n+e_m$ is in the convex hull of $B$.  Similarly, if $i,j,k,\ell$ are distinct, then $e_i+e_j+e_k+e_\ell\in[e_i+e_j,e_k+e_\ell]$ and thus is in the convex hull of $B$.  Continuing this process, we conclude that for any $N$, any sum of $2^N$ distinct $e_n$'s is in the convex hull of $B$.  Thus the convex hull is unbounded.
A: Consider the Chebyshev metric on $\Bbb R^3$, I claim that the only convex open set is empty set and the whole space (unbounded is sufficient to be a counterexample).
Let $U$ be an convex open set. Any non-empty open set would contain an open ball thus there are 2 elements share the same  y and z coordinate, say $\textbf x_1 = (x+\epsilon, y, z)$ $\textbf y_1 = (x-\epsilon, y, z)$.
Define $\textbf x_n = \textbf y_{n-1} +  (\epsilon,\epsilon,\epsilon)$ $\textbf y_n = \textbf x_{n-1} +  (-\epsilon,-\epsilon,\epsilon)$. You can verify that
$$d(\textbf x_{n-1}, \textbf x_n) + d(\textbf y_{n-1}, \textbf x_n) = d(\textbf x_{n-1}, \textbf y_{n-1}) = 2\epsilon$$
Therefore $U$ extends indefinitely to the $z$ direction. It is easy to show that it contains any element in the from $(x', y', z')$ where $x -\epsilon ≤x'≤ x + \epsilon $, same goes for $y'$ and $z'$ be arbitrary. Applying the same process can show that any element belong to $U$ thus $U = \Bbb R^3$.
