this theorem actually has two parts to it. for a domain $\Omega$ and any $f$ holomorphic on it,
by the cauchy integral theorem
\[\begin{align*} |f(z_0)| &= \left|\frac{1}{2\pi i} \int_{|\zeta-z|=R} \frac{f(\zeta)}{\zeta-z_0}d\zeta\right| \\ &= \left|\frac{1}{2\pi i} \int_{0}^{2\pi} \frac{f\left(z_0+Re^{i\theta}\right)}{(z_0+Re^{i\theta})-z_0} \cdot Rie^{i\theta} d\theta\right| \\ &= \left|\frac{1}{2\pi} \int_{0}^{2\pi} f\left(z_0+Re^{i\theta}\right) d\theta\right| \\ &\le \frac{1}{2\pi} \int_{0}^{2\pi} \left|f\left(z_0+Re^{i\theta}\right)\right| d\theta \\ &\le \frac{1}{2\pi} \int_{0}^{2\pi} \sup_{0 \le \theta \le 2\pi} \left|f\left(z_0+Re^{i\theta}\right)\right| d\theta \\ &= \frac{1}{2\pi} \left(2\pi \cdot \sup_{0 \le \theta \le 2\pi} \left|f\left(z_0+Re^{i\theta}\right)\right|\right) \\ &= \sup_{|\zeta-z_0|=R} |f(\zeta)| \end{align*}\]that bound on the integral comes from real analysis and also just plain calculus
ok last time i tried to prove this it went straight to hell, let's see if it's as bad as i remember
we have $|f(z_0)|=\sup_{|\zeta-z_0|=R} |f(\zeta)|$, so both of those inequalities that i put in that previous chain of expressions are actually equalities
in particular, this one here:
\[\frac{1}{2\pi} \int_{0}^{2\pi} \left|f\left(z_0+Re^{i\theta}\right)\right| d\theta =\frac{1}{2\pi} \int_{0}^{2\pi} \sup_{0 \le \theta \le 2\pi} \left|f\left(z_0+Re^{i\theta}\right)\right| d\theta\]by more real analysis this means that across all $\theta$, $|f(z_0)|=\left|f\left(z_0+Re^{i\theta}\right)\right|$
notice that this only gives us constant-ness wrt the magnitude though, not the actual number
for that, we have to pull out this deus ex machina:
\[g(z)=f(z) \cdot \overline{f(z_0)}\]and then use the cauchy integral theorem to evaluate it at $z_0$:
\[g(z_0)=\frac{1}{2\pi}\int_{0}^{2\pi} g\left(z_0+Re^{i\theta}\right)d\theta\]if this equality is true, surely it's still true if we take real parts on both sides:
\[\Re(g(z_0))=\frac{1}{2\pi}\int_{0}^{2\pi} \Re\left(g\left(z_0+Re^{i\theta}\right)\right)d\theta\]now we also have this inequality here:
\[\Re\left(g\left(z_0+Re^{i\theta}\right)\right) = \Re\left(f\left(z_0+Re^{i\theta}\right)\overline{f(z_0)}\right) \le \Re(g(z_0))\]where that last bound is because $|f(z_0)|=\left|f\left(z_0+Re^{i\theta}\right)\right|$ as derived earlier. with a fixed magnitude, the largest real part you can get when multiplying with $\overline{f(z_0)}$ only happens when you do it with $f(z_0)$
now notice that this inequality actually has to be met with equality, since if it were strict at any point then
\[\begin{align*} \frac{1}{2\pi}\int_{0}^{2\pi} \Re\left(g\left(z_0+Re^{i\theta}\right)\right)d\theta &< \frac{1}{2\pi}\int_{0}^{2\pi} \Re(g(z_0)) d\theta \\ &= \Re(g(z_0)) \end{align*}\]which would contradict our earlier equality.
so now we have $\Re\left(f\left(z_0+Re^{i\theta}\right)\overline{f(z_0)}\right) = \Re(g(z_0))$. combined with the fact that
\[\Re(g(z_0))=\Re(f(z_0)\overline{f(z_0)})=f(z_0)\overline{f(z_0)}\]and the constant-ness of the magnitude, this forces $f\left(z_0+Re^{i\theta}\right)=f(z_0)$ for all $\theta$
this pretty much proves it! all that's needed is to use that a convergent sequence with constant values implies universal constant-ness, which was proved back here $\square$