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.
By the Schwarz reflection principle, the formula $f(z)=\overline{f(1/\overline{z})}$ for $|z|>1$ extends $f$ analytically to the entire complex plane. Then by definition of the extension, $f(\mathbb{C})\subset f(\overline{\mathbb{D}}) \cup \overline{f(\overline{\mathbb{D}})}$, which is bounded by compactness.
Best Answer
Let $g (z)=\frac {1-if(z)} {1+if(z)}$. You can easily check that this maps $U$ into the open unit disk. It is holomorphic on $U$. Since it is bounded it has a removable singularity at each point of $S$. Hence it extends to a bounded entire function. By Louiville's Theorem it is a consatnt. This implies that $f$ is also a constant.