Suppose $U$ is a bounded open subset of the plane and $f$ is continuous on the closure of $U$ and holomorphic on $U$. Show that if $C \geq 0$ and $\begin{vmatrix} f(z) \end{vmatrix} \leq C$ for all $z$ on the boundary of $U$, then $\begin{vmatrix} f(z) \end{vmatrix} \leq C$ for $z \in U$.
This problem seems to scream Max Modulus Principle, but I am not sure how to use it here. I am not sure how, a non-constant function cannot attain its maximum is even applicable here.
Here would be my main argument, leaving the details and punchline for you:
Since $f$ is continuous on $\bar{U} = U \cup \partial U$, there must be a point $z_0 \in \bar{U}$ such that $|f(z_0)| = \max_{z \in \bar{U}}|f(z)|$ (since the continuous image of a compact set will be compact). If $z_0 \in U$, then $|f(z)| \leq |f(z_0)|$ for all $z \in U$. By MMP, this implies that $f$ is constant on $U$. Otherwise, $z_0 \in \partial U$.