Show that $f$ maps the entire unit disc onto itself. Suppose $f$ is analytic in the unit disc $D(0,1)$ and maps the unit circle into itself. Show then that $f$ maps the entire disc onto itself.
So the outline wants us to use the Max Modulus Theorem to show that $f$ maps $D(0,1)$ into itself. Then, use the fact that we proved that if $f:S \to T$, $f$ non-constant and analytic on $S$, and if $f(z)$ is a boundary point, $z$ is a boundary of $S$ to show that the mapping is onto.
I'm not sure if mapping the unit circle into itself means that $|f|=1$ on the unit circle. Also is the unit disc compact? Thanks!

Let $D$ denote $D(0,1)$, and $cl(D)$ the closed disk $\{|z| \leq 1\}$. The maximum modulus theorem immediately gives that $f$ maps $D$ into itself. So the issue is showing that it's onto. The following is a topological proof. 
By the open mapping theorem $f(D)$ is open, so $f(D) \cap D$ is open. Since $cl(D)$ is compact, so is $f(cl(D))$. Hence $f(cl(D))$ is closed. Hence $f(cl(D)) \cap D$ is a relatively closed subset of $D$. Since $f$ takes $\{|z| = 1\}$ to $\{|z| = 1\}$, $f(cl(D)) \cap D = f(D) \cap D$. Hence $f(D) \cap D$ is an open, closed, and nonempty subset of $D$. Thus by connectedness of $D$, you have that $f(D) \cap D = D$, which means $f$ is onto as well.
