The map $t\mapsto |f(z_0)|-|f(z_0+re^{it})|$ is non-negative, continuous, and its integral over $[0,2\pi]$ is $0$, hence for all $t\in [0,2\pi]$, we have that $|f(z_0+re^{it})|=|f(z_0)|$. It's true for $r$ small enough, hence the modulus of $f$ is constant on a ball centered on $z_0$. Using Cauchy-Riemann equations, it can be shown that $f$ is actually constant on the ball, and by connectedness on the domain.
i.e. I wanted to write a proof that:
$$\forall L \in \mathbb R \Big [\exists \epsilon \gt 0 \text{ s.t. } \forall N \in \mathbb R \ \exists x \in \mathbb R \text{ s.t. }\big (x \gt N \land \lvert \sin(x) - L \rvert \geq \epsilon \big) \Big]$$
We can do this using $\varepsilon = \frac{1}{2}$.
Case 1: $L > 0$
Here for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = -1.$$
Since $L > 0$ we have
\begin{align}
|\sin x - L|&= |-1 - L| \\
& = 1 + L \\
& > \frac{1}{2}
\end{align}
Case 2 $L < 0$
This case is nearly identical to the case for $L > 0$.
For any $N$ we can always find $x > N$ of the form $x = \frac{\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = 1.$$
Since $L < 0$ we have
\begin{align}
|\sin x - L|&= |1 - L| \\
& = 1 + |L| \\
& > \frac{1}{2}
\end{align}
Case 3 $L = 0$
This case can be dealt with immediately by re-using either of the $x$ from the previous two cases:
If $x = \frac{\pi}{2} + 2k\pi$, then
\begin{align}
|\sin x - L|&= |1 - 0| \\
& = 1 \\
& > \frac{1}{2}
\end{align}
Of course, this all depends on knowing the behavior of $\sin$: that $\sin$ wavers between $1$ and $-1$ "forever" and so on. Given this behavior, one would expect intuitively that the limit doesn't exist, i.e. $\sin x$ wouldn't "settle" upon some single value for large $x$.
Later in the book Spivak defines the trigonometric functions formally in a way that conforms with the geometric/unit circle versions with which you are probably already familiar. Continuity plays a big part in these later definitions.
Here though, the main thing is just the periodic nature of $\sin$.
Finally, you may be somewhat leery of statements like:
...for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
This has at its heart the idea that given any real number $N$ we can find a natural number $n$ with $n > N$. This fact is employed in a few places early on in the text, but isn't formally proven until Chapter 8 (3rd Ed.) "Least Upper Bounds".
Spivak cops to this "cheating" at the end of that chapter.
(As an aside, an earlier problem in Chapter 5, 5-12 gives a very useful result:
If $f(x) \leq g(x)$ for all $x$ (or even just for all $x$ in a neighborhood of $a$) then $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided these limits exist.
This can be shown to hold if we replace $a$ with $\infty$.
This, combined with your lemma that $|\sin x| \leq 1$ can be used to rule out your Case 1, i.e. since $-1 \leq \sin x \leq 1$ we must also have $-1 \leq \lim_{x \to \infty}\sin x \leq 1$, if that limit exists.)
Best Answer
I'd personally argue like this: