The lemma is on page 181.
Here's the lemma and its proof:
I don't understand how did they get:
$$\lim | S^{n_k+1}f(\omega)|\leq \gamma (\alpha - \epsilon)+(1-\gamma)\alpha$$
The term that is confusing is $\gamma (\alpha - \epsilon)$, I mean I understand why $\alpha - \epsilon$, but not why times $\gamma$.
I mean $P(\omega_k , F ) \geq \gamma$, and not less than or equal $\gamma$.
Since I do not have this book so what I know is the part of the book you showed in your question. So sorry for any misunderstanding. I think the reason is the following.
From (1.3) we know that $$|S^{n_k+1}f(\omega)| \leq \max_{\eta \in F}|S^{n_k}f(\eta)| P(\omega_k,F) + \max_{\eta \in F^c} |S^{n_k}f(\eta)|P(\omega_k, F^c).$$ Note that for any $\delta>0$, we can find $k$ such that $\max_{\eta \in F}|S^{n_k}f(\eta)| \leq \alpha-\epsilon+\delta$ and that $\max_{\eta \in F^c}|S^{n_k}f(\eta)| \leq \alpha+\delta$ so $$|S^{n_k+1}f(\omega)| \leq (\alpha-\epsilon+\delta) P(\omega_k,F) + (\alpha+\delta)P(\omega_k, F^c).\\ \leq \alpha+\delta-\epsilon P(\omega_k,F)\leq \alpha+\delta-\epsilon\gamma.$$ Then let $k\to 0$ so that $\delta\to 0$ we have the inequality.