For $z_1, z_2 \in \mathbb{B}^2$, define
$d(z_1, z_2) = \text{cosh}^{-1}(1+ \dfrac{2|z_1 - z_2|^2}{(1-|z_1|^2)(1-|z_2|^2)})$.
In my text book (Lee's Topological manifolds Problems 12-23), to prove the triangle inequality, it suffices to assume that one of the points is the origin. The following is the problem 12-23 in Lee's.
Prove the triangle inequality for the hyperbolic metric as follows. Show tha it suffices to assume tha one of the points is the orgin and use the identity $\text{cosh}^2 x - \text{sinh}^2 x = 1$ to show that $\text{sinh}d(z,0) = 2|z|/(1-|z|^2)$, and therefore by th Euclidean triangle inequality,
$$\text{cosh}d(z_1,z_2)\le \text{cosh}d(z_1,0)\text{cosh}d(z_2,0)+\text{sinh}d(z_1,0)\text{sinh}d(z_2,0)=\text{cosh}(d(z_1,0)+d(0,z_2)).$$
I don't understand why it is sufficient to assume that one of the points is the orgin. What should I do to extend it to general cases.
There is a subset of Möbius transformations that are isometries (those that map the disc to itself), and you can apply one to move one of the the points to the origin. (Specifically, $$ T(z) = \frac{z-a}{1-\bar{a}z} $$ maps the disc to itself and $T(a)=0$.)