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!
 A: If $w \in D$ and $w \not \in f(D)$ then
$$
z \mapsto \frac{1}{f(z)-w}
$$
is holomorphic on $\overline{D}$.  By the maximum modulus principle, for any $z \in D$
$$
\left|\frac{1}{f(z)-w}\right| \leq \max_{\omega \in \partial D} \left| \frac{1}{f(\omega)-w} \right| \leq \max_{\omega \in \partial D}\frac{1}{\left||f(\omega)|-|w|\right|} = \frac{1}{1-|w|}
$$
so $|f(z) - w| \geq 1-|w|$.  This means that $D \setminus f(D)$ is open.  By the open mapping theorem either $f$ is constant or $f(D)$ is open.  The latter implies that $f(D)=D$ since $D$ is connected.
Moreover, the image of a compact set under a continuous function is compact.  Therefore if $f$ is not constant then $f(\overline{D}) = \overline{D}$.
Or, alternatively, this bound shows that $w$ can be moved in a straight line towards $0$ while the radius of the "image free" disc around it increases.  For $w=0$ the bound becomes $|f(z)| \geq 1$ so that $f(D) \cap D = \emptyset$.  So either $f(D)$ contains all of $D$ or avoids it entirely.  In the latter case $f$ maps into the unit circle.  This would mean that $\overline{f} = f^{-1}$ but $\overline{f}$ can be holomorphic (complex differentiable) only if $f'$ vanishes identically.  The conclusion is that either $f(D)=D$ or $f$ is constant. 
A: 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.
A: A proof:
Suppose $f$ does not map $D^{2}$ to itself, then it must map a certain point $p\in D^{2}$ to the outside of $D^{2}$. Use the maximal modulus principle we may argue that $f$'s maximal value must be on the boundary of $S^{1}$. But this is impossible by our hypothesis. Thus $f$ must map $D^{2}$ to itself. 
You may question why $f$ must map a point in $D^{2}$ to the outside. In fact, $f$ cannot keeps the boundary while leaving any point in $D^{2}$ uncovered because $f$ is largely locally conformal(otherwise $f$ would be a constant function and the assertion would be trivial). Suppose $p$ is some point uncovered by $f$, and $d(p, f(D^{2})\not=0$, then $f$ would leave a 'hole' and this cannot hold if $f$ is conformal. The othercase $d(p,f(D^{2}))=0$ can be handled similarly. In the isolated points $f'=0$ the behavior is well controlled since $f$ is analytic. Thus $f$ must map at least one point from $D^{2}$ to $\mathbb{C}-D^{2}$ if $f$ is not an automorphism.
For your questions:
I'm not sure if mapping the unit circle into itself means that $|f|=1$ on the unit circle. 
Yes, this implies $|f(S^{1})|=1$. 
Also is the unit disc compact? 
This should be trivial. I guess your confusion stems from the fact that $f$ is only supposed to be analytical at the open disc (thus could have no analytical continuation on the boundary of the disc). But this does not violate the statement of the theorem, if you check it carefully. 
