I do not get part of the proof of the following theorem from Elias Wegert's Visual Complex Functions, page 181. Here, an essential singularity of an analytic function $f$ with Laurent series is defined to be an isolated singularity with Laurent series coefficients $c_k \neq 0$ for infinitely many negative $k$.
Theorem: If an isolated singularity $z_0$ of an analytic function $f$ is an essential singularity, then any neighborhood of $z_0$ intersects infinitely many isochromatic lines of the phase portrait of one and the same color, i.e. the level sets of a given argument.
Proof. Assume $f$ has an essential singularity at $z_0$. Let $D'$ be a punctured disk with center $z_0$. To begin with, we choose a point $z_1 ∈ D'$ with $f(z_1) \neq 0$ and an open disk $U_1$ centered at $z_1$ such that $U_1 ⊂ D'$ and $0 \not\in V_1 := f(U_1)$. By the open mapping principle, the set $V_1$ is open. In the next step we apply the Casorati-Weierstrass theorem to find a point $z_2 ∈ D' - U_1$ with $w_2 := f(z_2) ∈ V_1$. Further we choose an open disk $U_2$ centered at $z_2$ such that $U_2 ⊂ D' - U_1$ and $V_2 := f(U_2) ⊂ V_1$. Note that $V_2$ is an open set. Continuing in this manner, we get a sequence of pairwise disjoint disks $U_1,U_2,...,U_n ⊂ D'$ with images $V_k := f(U_k)$ forming a family of nested open sets $V_1 ⊃ V_2 ⊃ ...⊃ V_n$. Since $V_n$ is open and $f'$ has at most a countable number of zeros, we can choose $w^∗ ∈ V_n$ such that for $c := w^∗/|w^∗|$ the isochromatic set $S := \{z ∈ D' : f(z)/|f(z)| = c\}$ contains no zero of $f'$.
I get every part of the proof until here. Now the rest of the proof is what I do not get:
Consequently $S$ is the union of smooth isochromatic lines on which $|f|$ is strictly monotone. Because $w^∗$ belongs to $V_n$, and hence to all $V_k$, there exist $z^∗_k ∈ U_k$ with $f(z^∗_1)=f(z^∗_2)=...= f(z^∗_n)=w^∗$. This implies that all points $z_1,...,z_n$ must lie on different isochromatic lines with color $c$. Since $n$ can be chosen arbitrarily large this proves our claim.
Here is my question: why does the fact that $f' \neq 0$ on $S$ imply that $|f|$ is strictly monotone? I know the theorem that for a nonconstant analytic function $f$, the absolute value $|f|$ cannot have a maximum on an open set unless that maximum is $\infty$ and any minimum of $|f|$ must be zero. Also, why does the fact that "there exist $z^∗_k ∈ U_k$ with $f(z^∗_1)=f(z^∗_2)=...= f(z^∗_n)=w^∗$" imply that "$z_1,...,z_n$ must lie on different isochromatic lines with color $c$"? If I am not mistaken, shouldn't it be "$z^*_1,...,z^*_n $? Also, why must $z_1,...,z_n$ lie on different isochromatic lines of color $c$, i.e. why can't some of them be on a same isochromatic line?
I think there is a typo in your question that is a source of some confusion. Consulting my copy of the text, the proof contains the statement "Since $V_n$ is open and $f′$ has at most a countable number of zeros, we can choose $w^∗ ∈ V_n$ such that for $c := w^∗/|w^∗|$ the isochromatic set $S := \{z ∈ D ̇ : f(z)/|f(z)| = c\}$ contains no zero of $f′$. "
This condition on the derivative $f'$ (and a similar condition on absence of zeros of $f$) implies that $w=\ln f(z)$ is free of critical points, so the level sets of the real and imaginary parts of $w=\ln f(z)$ are smooth curves. (This can be deduced by the real-variable implicit function theorem of multivariable calculus.) These level sets of the imaginary part of $\ln f$ are what the author calls isochromatic lines.
Additional edited comment. And the reason that $|f|$ is monotone on the isochromatic line is that $\ln |f|$ is monotone. And that is because $w= u+iv = \ln f(z)$ has the property that $u=\ln |f|$ is monotone on the isochromatic level curve where $v= Arg (f)$ is constant. In fact the gradient vectors of the functions $u$ and $v$ are perpendicular in the $z$ plane, by the Cauchy-Riemann equations, so locally we see that as we travel on each level curve of $v$ we are traveling along the direction of steepest ascent (or descent ) for $u$. This local monotonicity property can be patched together to deduce global monotonicity as you travel along the isochromatic line.