Bloch's Theorem. Let $f$ be an analytic function on a region containing the closure of the disk $D= \lbrace z ; |z|<1 \rbrace$ and satisfying $f(0)=0$ and $f'(0)=1$, then there is a disk $S\subset D$ on which $f$ is one-one and such that $f(S)$ contains a disk of radius $1/72$.
Proof. Let $K(r)=\max \lbrace|f'(z)| ; |z|=r\rbrace$ and let $h(r)=(1−r)K(r)$. It is easy to see that $h:[0,1]⟶\mathbb{R}$ is continuous, $h(0)=1$, $h(1)=0$. Let $r_{0}=\sup \lbrace r ; h(r)=1\rbrace$; then $h(r_{0})=1$, $r_{0}<1$ and $h(r)<1$, if $r>r_{0}$.
Why do we have that $h(r)<1$, if $r>r_{0}$?
How can we show that $h$ is continuous ?
Suppose $r >r_0$ and $h(r) >1$. Look at $h$ on $[r,1]$. Since $h(r)>1$ and $h(1)=0$ there exists $s \in (r,1]$ such that $h(s)=1$. But this contradicts the definition of $r_0$.
Continuity of $h$ follows from the fact that $K(r)=\sup \{|f'(z)|: |z| \leq r\}$ and the fact that $f'$ is uniformly continuous on closed balls.