= Solution
Fix $z\in K$. If $z\in\partial K$, the conclusion is immediate. Otherwise $z$ is in the <interior> of $K$. If $f(z)=0$, nonnegativity of the <boundary> <supremum> suffices. For $f(z)\ne0$, the complex <Hahn-Banach theorem> gives a <bounded linear functional> $\ell:A\to\mathbb C$ with $\|\ell\|=1$ and $\ell(f(z))=\|f(z)\|$. The scalar function $g=\ell\circ f$ is <holomorphic>.
Let $V$ be the <connected component> of the <interior> of $K$ containing $z$. It is bounded, its closure is contained in $K$, and $\partial V\subseteq\partial K$. To see the last inclusion, a <boundary> point of $V$ lying in the <interior> of $K$ would have a small <connected> open ball in that <interior> meeting $V$; the ball would belong to the same component, contradicting that it is a <boundary> point. The scalar <maximum modulus principle on a bounded domain> therefore gives
$$
\|f(z)\|=|g(z)|\le\sup_{w\in\partial V}|g(w)|\le\sup_{w\in\partial K}\|f(w)\|.
$$
For <completeness>, this <boundary> form of the <maximum modulus principle> follows by taking the maximum on the <compact> closure of $V$. An <interior> maximum makes $g$ constant on $V$, in which case <continuity> gives the same value on its nonempty <boundary>. Thus the <norm maximum principle for a Banach-space-valued holomorphic function> applies to arbitrary <compact> $K$, even when it is disconnected or has nonsmooth <boundary>:
$$
\boxed{\|f(z)\|\le\sup_{w\in\partial K}\|f(w)\|.}
$$
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