Suppose the following conditions hold true:
(1) The function $f$ is analytic and contractive in the open unit disk.
(2) $f(0)=0$.
(3) $\exists z_1 \ne z_2 \in B(0,1)$, $|z_1|=|z_2|$, $f(z_1)=f(z_2)$.
Prove that $|f(z_1)|=|f(z_2)| \leqslant |z_1|^2=|z_2|^2$.
Let $f(z_1)=f(z_2)=y$; by hypothesis we know $|f(z)| <1, z \in \mathbb D$ (and actually $|f(z)| \le |z|$ since $f(0)=0$ but that would help directly only if we knew that $f'(0)=0$) so in general we proceed as follows:
Let $g(z)=\frac{f(z)-y}{1-\bar y f(z)}:\frac{(z-z_1)(z-z_2)}{(1-\bar z_1 z)(1-\bar z_2 z)}$
Since $f(z_1)=f(z_2)=y$, it follows that $g$ is also analytic in the unit disc and since $|f(z)| < 1$ the first term $|\frac{f(z)-y}{1-\bar y f(z)}| \le 1$, while $|\frac{(z-z_1)(z-z_2)}{(1-\bar z_1 z)(1-\bar z_2 z)}| \to 1 , |z| \to 1$ so by maximum modulus $|g(z)| \le 1, |z| <1$ by the usual methods (for every $\epsilon >0$ there is $r(\epsilon), |\frac{(z-z_1)(z-z_2)}{(1-\bar z_1 z)(1-\bar z_2 z)}| \ge 1-\epsilon, 1> r=|z| > r(\epsilon)$ etc)
This means $|g(0)| \le 1$ so using finally that $f(0)=0$ we get:
$|f(z_1)|=|y| \le |z_1||z_2|=|z_1|^2$ so we are done!