This is not true.
Let $K$ be, say, the unit ball in a Hilbert space, equipped with its weak topology. Let $\xi$ be a norm-one linear functional, and consider $U := K \cap \{\xi \in [-1, -c) \cup (c, 1]\}$ for $0 < c < 1$. It is an open set, however, its convex hull is $\{x \in K \, | \, \Vert \mathrm{pr}_{\xi^\perp} x \Vert < \sqrt{1 - c^2}\}$, where $\mathrm{pr}_{\xi^\perp}$ is the orthogonal projection on the hyperplane $\xi^\perp$. This set is clearly not open in the weak topology.
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
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.