Topology basis consisting of convex sets in metric spaces

Question

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.

Answer 1

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.