$f$ be non constant analytic on the closed unit disk,$|f|=1$ if $|z|=1$,we need to show image of $f$ contains the unit disk.
My thoughts:
whenever $|\omega|<1$ if I show that $g(z)=f(z)-\omega$ has a root in $D$ the we are done
On $|z|=1$ $|f(z)-g(z)|=|\omega|<1=|f(z)|$ Hence by Rouches Theorem $f(z)$ has a root in $D$
Best Answer
Assume $f$ does not have a zero. Applying the maximum modulus principle to $1/f(z)$ gives a contradiction ($f$ constant, contradicting the hypothesis), so we find that any analytic function that has modulus $1$ on $|z|=1$ has a zero.
Fix $a$ in the open unit disk. Consider the Blaschke factor switching $0$ and $a$. Composing $f(z)$ with this factor gives another analytic function with modulus $1$ on the unit circle, so the composition has a zero in the unit disk. But this means $f(z)$ takes on the value $a$ in the unit disk.