In Saff's Complex Analysis book, the following OMT is stated.
If $f$ is non-constant and analytic in a domain $D$, then its range $f(D) := \{ w ~|~ w = f(z)~ \text{for some}~ z \in D \}$ is an open set.
Following is the outline of the proof.
Suppose $f (z)$ is analytic in some open neighborhood of a point $z_0$ and $f (z_0) = 0$. Since zeros of $f$ are isolated, there is some circle $C$, centered at $z_0$ and lying in the neighborhood, such that $f$ is non-zero on $C$. Let $\sigma := \min_{z \in C} |f(z)|$ and consider $g(z) = f(z) -c$, where $c \in \mathbb{C}$ such that $|c| < \sigma$. Since $|g(z)-f(z)| < |f(z)|$ on $C$, it follows from the Rouch'{e}'s theorem that $f (z) - c$ also achieves the value $0$, and hence $f$ achieves the value $c$, inside the circle.
I couldn't understand the rest of the argument of the proof. Any help in understanding this is much appreciated.
It is desired to prove that $f(D)$ is open in $\mathbb C$. That is, for any $z\in D$, there exists $r_z>0$ such that the ball $B(f(z), r_z)$ centered at $f(z)$ with radius $r_z>0$ is a subset of $f(D)$.
So take any $z\in D$. WLOG assume that $f(z)=0$. The first part in OP shows that $B(f(z),\sigma)\subset f(D)$. This is because if $w\in B(f(z),\sigma)$, then $|w-f(z)|=|w-0|<\sigma\implies |w|<\sigma$ and by the first part in OP, $f(u)-w=0$ can be solved for some $u\in D$. In other words, $w\in f(D)$. This proves that $B(f(z),\sigma)\subset f(D)$.
If it is not true that $f(z)=0$, then you take $F(u): =f(u)-f(z)$ so that $F(z)=0$. Then, use the above idea on $F$.