On page 129 of Conway's Functions of one complex variable, there is a version of generalized Maximum Modulus Theorem (let $\partial_{\infty}G$ denote the boundary of $G$ in $\mathbb C_{\infty}$ and $\bar G_{\infty}$ be the closure in $\mathbb C_{\infty}$):
Theorem 1.4 Let $G$ be a region in $\mathbb C$ and $f$ an analytic function on $G$. Suppose there is a constant M such that $\limsup_{z\to a}|f(z)|\leq M$ for all $a\in \partial_{\infty}G$. Then $|f(z)|\leq M$ for all $z\in G$.
Conway's proof is somewhat tedious, and I think a more straightforward solution is as follows:
First we let $g(a)=\limsup_{z\to a}|f(z)|$ from $\bar G_{\infty}$ to $[0,\infty]$, then $g$ is a upper semi-continuous function: for any $a$ and $\epsilon$, there is a neighborhood $U_a$ s.t. $\sup_{U_a}|f|<g(a)+\epsilon$, but then for $b\in U_a$, $g(b)=\limsup_{z\to b}|f(z)|\leq \sup_{U_a}|f|$, hence $g(b)\leq g(a)+\epsilon$ and $\limsup_{b\to a}g(b)\leq g(a)$.
Second, since $\mathbb C_{\infty}$ is compact, $\bar G_{\infty}$ is also compact, whence the extreme value theorem for semi-continuous function applies. Say $g$ assumes its maximum at some point $a$, if $a\in \partial_{\infty}G$, and note that $g\big|_{G}=f$, we obtain $f(z)\leq\sup_{\partial_{\infty}G}g$ in $G$. If $a\in G$, then the component of $G$ (say $H$) that contains $a$ must be constant, hence for all $z\in\partial H, g(z)=g(a)$, but $\partial H\in\partial_{\infty}G$, and we're done.
My question: (1) Is the proof above correct?
(2)In the proof above, we donnot need $G$ to be connected, hence the condition that $G$ is a region can be replaced by $G$ is any open set?