Here is a counterexample: no non-empty open set in this metric space is convex in your sense.
Let $D=\{0,1\}$ with the discrete topology, and let $X=D^{\Bbb N}$, the Cartesian product of countably infinitely many copies of $D$, so that the elements of $X$ are the sequences $x=\langle x_n:n\in\Bbb N\rangle$ such that each $x_n$ is either $0$ or $1$. (This space is homeomorphic to the well-known middle-thirds Cantor set.)
For distinct $x,y\in X$ let $\delta(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}$, the first index at which $x$ and $y$ disagree; then
$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
2^{-\delta(x,y)},&\text{otherwise}
\end{cases}$$
is a metric on $X$. Suppose that $x,y$, and $z$ are distinct points of $X$ and that $d(x,y)+d(y,z)=d(x,z)$; then
$$2^{-\delta(x,y)}+2^{-\delta(y,z)}=2^{-\delta(x,z)}\,.\tag{1}$$
Let $n=\max\{\delta(x,y),\delta(y,z),\delta(x,z)\}$; clearly $\delta(x,y),\delta(y,z)>\delta(x,z)$, so we may assume that $n=\delta(x,y)$. Let $k=\delta(y,z)\le n$ and $\ell=\delta(x,z)<k$. Then we can rewrite $(1)$ as $2^{-n}+2^{-k}=2^{-\ell}$, and after dividing through by $2^{-n}$ we have
$$1+2^{n-k}=2^{n-\ell}\,.$$
Clearly this is possible if and only if $2^{n-k}=1$ and $2^{n-\ell}=2$, i.e., iff $k=n$ and $\ell=n-1$. Since $k=n$, we know that $x_i=y_i=z_i$ for $i<n$ and $x_n\ne y_n\ne z_n$. But then $x_n=z_n$, so $x_i=z_i$ for $i\le n$, and $n-1=\ell=\delta(x,z)\ge n+1$, which is absurd.
Thus, for $x,y,z\in X$ the relation $d(x,y)+d(y,z)=d(x,z)$ holds iff $y\in\{x,z\}$, so the only convex sets in $X$ are the singletons and doubletons. Every non-empty open set in $X$ is infinite (and in fact has cardinality $2^\omega=\mathfrak{c}$), so $\langle X,d\rangle$ does not have a base of convex sets.
Best Answer
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.