I am trying to prove the following: Let be $f: \mathbb{D} \mapsto \mathbb{D}$, an analytic function. If for $z,w \in \mathbb{D}$, $z\neq w$, we have
$$ \left| \frac{f(z)-f(w)}{1- \overline{f(w)}f(z)} \right| = \left| \frac{z-w}{1 - \overline{w}z} \right|, $$ then we have this equality for any pair of distinct points in the disk.
This my work so far:
I consider the following function:
$$ \varphi (\zeta) = B_{f(w)} \circ f \circ B_w (\zeta) $$ where $B_w(\zeta) := \frac{w - \zeta}{1-\overline{w}\zeta},~ \text{and}~ \tau_{f(w)} (\zeta) := \frac{f(w) - \zeta}{1-\overline{f(w)}\zeta}, ~\zeta \in \mathbb{D} $. Those fucntions are Blaschke factors. $\varphi$ is clearly analytic, mapping the disk on itslef, and $\varphi (0)=0$. Also, if we suppose $z_0=B_w(z)$, we have: \begin{align*} |\varphi(z_0)| &= \left| B_{f(w)} \circ f \circ B_w (z_0) \right|\\ &=\left| B_{f(w)} \circ f \circ B_w \circ B_w(z) \right| \\ &=\left| B_{f(w)} \circ f(z) \right| ~~~\text{(because $B_w \circ B_w =id)$} \\ &=\left| \frac{f(z)-f(w)}{1- \overline{f(w)}f(z)} \right|\\ &= \left| \frac{z-w}{1 - \overline{w}z} \right| ~~\text{(hypothesis)}\\ &=|z_0|, \end{align*} we can then apply $\textit{Schwarz Lemma}$ and say that $\varphi (\zeta)=\gamma z$ where $|\gamma|=1$. Taking $\zeta=B_w(a)$, for some $a \in \mathbb{D}$ we obtain the following: \begin{align*} |\varphi(B_w(a))|&=\left| \frac{f(a)-f(w)}{1- \overline{f(w)}f(a)} \right|\\ &=|\gamma B_w(a)|\\ &=|\gamma| \left| \frac{a - w}{1-\overline{w}a} \right|\\ &=\left| \frac{a - w}{1-\overline{w}a} \right|. \end{align*}
Here I proved, if i'm not wrong, that the equality works for any $z \in \mathbb{D}$ but it only works with the $w$ we started with. Any ideas how to prove that it should work for any pairs of complex numbers in the disk? Thank you!
For this problem you have to know that the Poincaré distance between any two points $z$, $w\in D$ is given by $$d(z,w)={\rm artanh} \left|{z-w\over 1-\bar wz}\right|\ ,$$ and that this distance is invariant under conformal automorphisms of $D$ (called Blaschke factors in your question). In your case you are told that a certain holomorphic function $f:\>D\to D$ satisfies $$d\bigl(f(z),f(w)\bigr)=d(z,w)\tag{1}$$ for a certain pair of different points $z$, $w\in D$, and it is claimed that such an $f$ automatically is a conformal automorphism of $D$.
That's where the Schwarz Lemma comes in: It is about the special case $w=f(w)=0$, and indeed says that $(1)$ for a single $z\ne0$ guarantees $f$ to be an automorphism of $D$. But any $f$ satisfying your assumption can be conjugated to a such a special $f$ by Blaschke factors (that's what you have done in your calculations). The rest is "going through the motions".