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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 $f(z)\ne 0$, the complex Hahn-Banach theorem gives a bounded linear functional $\ell :A\to \mathbb{C}$ with $\Vert \ell \Vert =1$ and $\ell (f(z))=\Vert f(z)\Vert$. 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 givesFor 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:

For a Banach algebra-valued holomorphic function, set $p_n(w)=\Vert f(w{)}^{2^n}{\Vert }^{1/2^n}$. These continuous functions decrease to the spectral radius by the spectral radius formula. If $M={\sup }_{\partial K}r(f(w))$, the continuous functions $\max \{p_n,M\}$ on $\partial K$ decrease to the continuous constant $M$. The Dini theorem gives uniform convergence. Apply the norm maximum principle for a Banach-space-valued holomorphic function to each power, and then let $n\to \infty$. This proof does not assume continuity of spectral radius.