I am trying to understand a proof of this theorem:
Suppose $f$ is holomorphic in open set that contains the closure of a disk D. If C denotes the boundary circle of this disk with positive orientation,then $$f(z)=\frac{1}{2\pi}\int_C\frac{f(\zeta)}{\zeta - z}d\zeta$$ for any point $z \in D$.
Now they have chosen the function $F(\zeta)=\frac{f(\zeta)}{\zeta - z}$ and have chosen a keyhole and we have
$\displaystyle \int_{\Gamma_{\delta,\epsilon}}F(\zeta)d\zeta=0$ by Cauchy's theorem.
They call the small circle $C_\epsilon$
because it has $\epsilon$ radius.
However, later the authors split $F(\zeta)$ as $F(\zeta)=\frac{f(\zeta)-f(z)}{\zeta - z}+\frac{f(z)}{\zeta - z}$
and they comment that as $f$ is holomorphic, the term $\frac{f(\zeta)-f(z)}{\zeta - z}$ is bounded and its integral over
$C_{\epsilon}$ goes to $0$ as $\epsilon \to 0$.
I really don't understand that comment and I shall appreciate if someone could please tell me why the comment implies that the integral of $\frac{f(\zeta)-f(z)}{\zeta - z}$ over $C_{\epsilon}$ is $0$.
That is just Cauchy's Estimation Lemma or the M-L inequality, which gives
$$\left|\int_{C_\epsilon}\frac{f(\zeta)-f(z)}{\zeta-z}d\zeta\right|\le M2\pi\epsilon\xrightarrow[\epsilon\to0]{}0$$
Continuity of the integrand is important here and you missed it, perhaps.