$f(z)\ne 0$ if $|z|<1/2$ in this situation?

Question

Assumptions.

1. $D$ is the open unit disk $\{z\in \mathbb C : |z|<1\}$.
2. $f$ is holomorphic in $D$.
3. $f(0)=1/2$.
4. $\sup \{ |f(z)| : z \in D \} =1$.

Question. $f(z)\ne 0$ if $|z|<1/2$?

Usually I rely on Rouché's theorem to count the number of zero of a holomorphic function. But in this case I don't see how to apply the theorem. I tried to find a holomorphic function $g$ in $D$ such that $$ \text{If $|z|<\frac{1}{2}$, then $g(z)\ne 0$.} $$ and at the same time $$ \text{ $|f(z)-g(z)|<|g(z)|$ if $|z|=\frac{1}{2}$. } $$ But it was not successful because I don't know much about $f$. Maybe this is not the way. Is there any way to apply Rouché's theorem?

Thanks for any answer, no matter Rouché's theorem is used or not.