Clarification on one-to-one mapping of a disk in complex analysis

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I have a some difficulties understanding the proof in section 7.2 of "Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics" by Edward Saff and Arnold Snider.

Theorem 1 states:

"If $f$ is analytic at $z_0$ and $f'(z_0) \neq 0$, then there exists an open disk $D$ centered at $z_0$ such that $f$ is one-to-one on $D$."

The provided proof sketch involves establishing an inequality $|f(z_1) - f(z_2)| \geq |\frac{f'(z_0)}{2}||z_2 - z_1|$ for points $z_1, z_2$ within this disk $D$. However, I'm struggling to understand how this inequality holds.

Specifically, my questions are:

Are all points $z_1$ and $z_2$ considered to be within the same open disk $D$ centered at $z_0$? Should I interpret $f(z)$ around a point $z$ on the disk $D$ as the power series expansion about $z$? How do the moduli $|f(z_1) - f(z_2)|$ and $|f'(z_0)(z_2 - z_1)|$ relate to the radius of convergence of the power series expansion about individual points $z_1$ and $z_2$? I'm referring to the proof of Theorem 1 presented in the book in chapter 7.2.

Any clarification or insight into these questions would be greatly appreciated. Thank you!