Is there a simple way to prove the following inequality without using Möbius transformations:
$\left| \frac{{z}_{1}-{z}_{3}}{1-{z}_{3}\bar{{z}_{1}}} \right| \leq \left| \frac{{z}_{1}-{z}_{2}}{1-{z}_{2}\bar{{z}_{1}}} \right| + \left| \frac{{z}_{2}-{z}_{3}}{1-{z}_{3}\bar{{z}_{2}}} \right|$,
where ${z}_{i}$ is a complex number and $\left|{z}_{i}\right|<1$ holds for $i=1,2,3$
(This would be my approach if I were to prove the inequality by using a Möbius Transformation: Use the hyperbolic metric $d({z}_{1},z_{2})=2tanh^{-1}\left| \frac{{z}_{1}-{z}_{2}}{1-{z}_{2}\bar{{z}_{1}}} \right|$. Apply the Möbius Transformation $\phi(x)=\frac{{z}_{1}-x}{1-x\bar{{z}_{1}}}$ so now we have $d(0,\phi(z_{2}))=d({z}_{1},z_{2})$. By combining this result and the fact that $d$ is a metric (so the triangle inequality holds) and tanh is bijective it should be easy to prove the above inequality.)
This is the completion of your idea of using the hyperbolic distance function $$ d(z_1, z_2)=2\tanh^{-1}\left\lvert \frac{z_1-z_2}{1-z_2\overline{z_1}}\right\rvert.$$
To begin, we note that the hyperbolic arctangent is superadditive for nonnegative arguments; $$ \tanh^{-1}(x+y)\ge \tanh^{-1}(x)+\tanh^{-1}(y),\qquad \forall x, y\in[0, 1).$$ (This follows from the fact that $\tanh^{-1}$ is convex, increasing, and that $\tanh^{-1}(0)=0$; see this answer). Combining the triangle inequality for $d$ and this superadditivity, we have that $$ \begin{split} \tanh^{-1}\left\lvert \frac{z_1-z_2}{1-z_2\overline{z_1}}\right\rvert&\le\tanh^{-1}\left\lvert \frac{z_1-z_3}{1-z_3\overline{z_1}}\right\rvert+\tanh^{-1}\left\lvert \frac{z_3-z_2}{1-z_2\overline{z_3}}\right\rvert \\ &\le\tanh^{-1}\left(\left\lvert \frac{z_1-z_3}{1-z_3\overline{z_1}}\right\rvert + \left\lvert \frac{z_3-z_2}{1-z_2\overline{z_3}}\right\rvert\right), \end{split} $$ hence the claim, since $\tanh^{-1}$ is increasing.