Let $f$ be holomorphic in a domain $U$. Let $a\in U$. Show there cannot exist $\varepsilon > 0$ such that $D(a,\varepsilon) \subset U$ and $|f(z)|<|f(a)|$ for all $z\in D(a,\varepsilon)$ (i.e., $|f|$ cannot have a strict local maximum).
Here's my proof, which I get this feeling is missing something as it's rather short: (Note by $\bar D$ I mean the closed disk).
With $\varepsilon>0$ given, take $0<\delta<\varepsilon$. Then $\bar D(a,\delta) \subset D(a,\varepsilon)$. $f$ is holomorphic here, so by Cauchy's integral formula; $$f(a)=\dfrac{1}{2\pi i}\displaystyle\int_{\partial\bar D} \dfrac{f(z)}{z-a}\ dz$$
It follows that $$|f(a)|=\dfrac{1}{2\pi }\left|\displaystyle\int_{\partial\bar D} \dfrac{f(z)}{z-a}\ dz\right|$$
We use the estimation lemma on the integral; $L(\partial\bar D) = 2\pi\delta$, and for an upper bound; $$|\dfrac{f(z)}{z-a}|=\dfrac{|f(z)|}{\delta} < \dfrac{|f(a)|}{\delta}$$
Putting together, $$|f(a)|<\dfrac{1}{2\pi}2\pi\delta\dfrac{|f(a)|}{\delta} = |f(a)|$$
Where we have the strict inequality as our $M$ is strictly greater than the integrand. Of course $|f(a)|<|f(a)|$ is a contradiction. Since $0<\delta<\varepsilon$ were arbitrary, this holds for every open disc in $U$ centred at $a$.
Does this proof hold, or is my suspicion justified and I've fallen prey to some caveat?