Let $\Sigma$ be a collection of holomorphic, one-to-one function from some simply connected region $\Omega$, which map $\Omega$ into the open unit disc $U$.
Fix $z_0 \in \Omega$ and put $$\eta = \sup\{|\psi'(z_0)|:\psi \in \Sigma\}$$ We want to find a function $h \in \Sigma$ such that $h'(z_0) = \eta$.
Now, before doing that, shouldn't we first make sure that $\eta < \infty$? But I don't know how to prove it. Of course $|\psi'(z_0)| < \infty$ for every $\psi \in \Sigma$ since they are holomorphic,but that does not suffice, does it?
Can somebody help me clear out this point?
P.S. As someone may have recognized,it's the final part of the proof of the Riemann Mapping theorem from "Real and Complex Analysis", by Rudin
$|h(z)| < 1$ for all $h \in \Sigma$ and all $z \in \Omega$, so $\Sigma$ is uniformly bounded.
Montel's theorem states that a uniformly bounded family is normal, i.e. every sequence of members of $\Sigma$ contains a subsequence which converges uniformly on compact subsets of $\Omega$. (Theorem 14.6 in my edition of Rudin.)
Now from the definition of $\eta$ it follows that there is a sequence $(h_n)$ in $\Sigma$ such that $|h_n'(z_0)| \to \eta $. Since $\Sigma$ is normal, this sequence has a subsequence $(h_{n_k})$ which converges uniformly on compact subsets to a holomorphic function $H$.
Then $h_{n_k}' \to H'$ uniformly on compact subsets of $\Omega$, and in particular $|h_{n_k}'(z_0)| \to |H'(z_0|$ and therefore $\eta = |H(z_0| <\infty $.
(The proof in Rudin's book then continues to show that this $H$ actually is a conformal mapping from $\Omega$ onto the unit disk.)