= Solution
The closed <unit disk> is a <compact set>, and $|f|$ is <continuous> on it because $f$ is <holomorphic> on an <open neighbourhood>. The <extreme value theorem> therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; <continuity> then gives the same value on the boundary. Consequently
$$
\boxed{\max_{|z|\leq1}|f(z)|=\max_{|z|=1}|f(z)|.}
$$
The assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.
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