Sorry for the late response. Here's a proof of the open mapping theorem assuming the maximum modulus principle.
First, we need the "minimum modulus principle". That is, if $f$ is a non-constant analytic function on an open connected set $D \subset \mathbb{C}$, and $f$ has no zeroes in $D$, then $| f |$ cannot attain a minimum in $D$. The proof follows trivially by applying the maximum modulus principal to the function $1/f$ which is analytic on $D$.
Now suppose $D \subset \mathbb{C}$ is open and connected, and $f$ is a non-constant analytic function on $D$. Let $U \subset D$ be open, and let $w_0 \in f(U)$, say $w_0=f(z_0)$ with $z_0 \in U$. We must show that there is a disc centered at $w_0$ which is contained in $f(U)$.
Choose $t>0$ so that $\overline {D_t(z_0)} \subset U$ and $f(z) \neq w_0$ for any $z \in \overline {D_t(z_0)}$ other than $z_0$. Let $m=\inf \{|f(z)-w_0| : |z-z_0|=t \} > 0 $. Suppose $|w-w_0| < m/3$, and that there is no $z \in U$ such that $f(z)=w$. Then the function $g(z)=f(z)-w$ is analytic, non-constant, and has no zeroes in the open connected set $D_t(z_0)$, so the minimum modulus principle shows that $g$ cannot attain a minimum modulus in $D_t(z_0)$. However, $g$ does attain a minimum modulus in the compact set $ \overline {D_t(z_0)} $, so this minimum modulus must occur on the boundary circle defined by $|z-z_0|=t$. But if $|z-z_0|=t$, then
$|g(z)|=|f(z)-w| \geq |f(z)-w_0| - |w_0-w| \geq 2m/3 $, and
$|g(z_0)| = |w_0-w| < m/3 < 2m/3$.
This gives a contradiction since $z_0$ is obviously in the interior of the disc in question. Therefore $f(z)=w$ for some $z \in D_t(z_0)$, and $D_{m/3}(w) \subset f(U)$, showing that $f(U)$ is open and proving the theorem.
Suppose $\operatorname{re} f(z) \ge \alpha$ and let
$g(z) = {1 \over f(z)-\alpha+1 }$. Since $\operatorname{re} (f(z)-\alpha+1) \ge 1$, we have $|g(z)| \le 1$ for all $z$ hence $g$ is a (non zero) constant. Since
$f(z) = \alpha-1+{1 \over g(z)}$, we see that $f$ is constant.
If $\operatorname{re} f(z) \le \alpha$, then repeat the above with $-f$ (and
$-\alpha$, of course).
Best Answer
I think it was James Littlewood who wrote "Could a fellowship be awarded for a dissertation of 2 lines? Presumably. For example:
Theorem. A non-constant complex polynomial $p$ has a zero.
Proof. Otherwise $1/p$ is a non-constant bounded entire function." (END QUOTE).
The Fundamental Theorem of Algebra was actually first proved by Gauss, well before Liouville's theorem (and Littlewood knew this). But it can be easily proved using Liouville's result:
Let $p:\Bbb C\to \Bbb C$ be a non-constant polynomial. Since $|p(z)|\to \infty$ uniformly as $|z|\to \infty$ there exists $r\in \Bbb R^+$ such that $|z|\geq r\implies |p(z)|>|p(0)|.$ Therefore $\inf \{|p(z)|:z\in \Bbb C\}=\inf \{|p(z)|:|z|\leq r\}=$ $=\min \{|p(z)|:|z|\leq r\}.$
This $\min$ must be $0,$ otherwise $1/p$ is a non-constant bounded entire function.