I understand the proof before the circled lines, but why do we want the roots of the equation $f(z)=a$ to be simple for $a\not = w_0$? Why do we know we can choose $\epsilon$ so that $f'(z)$ does not vanish for $0<|z-z_0|<\epsilon$? I think the proof of the theorem is finished before the circled lines.
A corollary after this theorem is that a nonconstant analytic function maps open sets onto open sets.
We have $f(z) = (z-z_0)^ng(z)$ with an analytic function $g$ such that $g(z_0) \ne 0$. Then $f'(z) = n(z-z_0)^{n-1}g(z) + (z-z_0)^ng'(z) = (z-z_0)^{n-1}(ng(z) + (z-z_0)g'(z))$. Thus it suffices to show that we can choose $\varepsilon$ such that $ng(z) + (z-z_0)g'(z) \ne 0$ for $\lvert z - z_0 \rvert < \varepsilon$. Let $\lvert g(z_0) \rvert = r > 0$ and $\lvert g'(z_0) \rvert = s$. We can choose $\varepsilon < \frac{r}{2(s+1)}$ such that $\lvert ng(z) \rvert > \frac{r}{2}$ and $\lvert g'(z) \rvert < s + 1$ for $\lvert z -z_0 \rvert < \varepsilon$. For $\lvert z -z_0 \rvert < \varepsilon$ we therefore get $$\lvert ng(z) + (z-z_0)g'(z) \rvert \ge \lvert ng(z) \rvert - \lvert (z-z_0)g'(z) \rvert = \lvert ng(z) \rvert - \lvert z-z_0 \rvert \lvert g'(z) \rvert >\\ \frac{r}{2} - \frac{r}{2(s+1)}(s+1) = 0,$$ that is $ng(z) + (z-z_0)g'(z) \ne 0$.
But if we know that $f'(z) \ne 0$, then $h(z) = f(z)-a$ can have no zero of order $>1$ at $z$ (otherwise $h'(z) = f'(z) = 0$).