Let $\mathbb{S}^{1} = \{ z \in \mathbb{C} : |z| = 1 \}$ and $f: \mathbb{S}^1 \longrightarrow \mathbb{C}$, $f \in \mathcal{C}^{0}(\mathbb{S}^{1})$, i.e., $f$ is continuous. Define for $z \in D_1(0) = \{z \in \mathbb{C} : |z| < 1 \}$, $$ F(z) = \frac{1}{2\pi i}\int_{\mathbb{S}^{1}} \frac{f(\zeta)}{\zeta - z}d\zeta$$
I'd like to proof $F \in H(D_1(0))$, Holomorphic functions in $D_1(0)$. I've started by definition, for some $z_0 \neq z$, we have the Newton's quotient $$ \frac{F(z) - F(z_0)}{z - z_0}$$
After calculations, we see $$\frac{F(z) - F(z_0)}{z - z_0} = \frac{1}{2\pi i} \int_{\mathbb{S}^{1}} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)}d\zeta$$
My idea is taking the limit, but I believe that is wrong, because I don't know nothing about of $f$, and if I can change integral and limits. Do we need more information about $f$? Like a uniform convergence of some sequence of $(f_n)_{n \in \mathbb{N}}$, $f_n(\zeta) \rightarrow \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)}$.
Thank you.
You are almost done! Recall that since you only care about the limiting behaviour of $z$, and since $z_0$ lies 'far' from $\mathbb{S^1}$, so $z$ must also be away from $\mathbb{S^1}$ after some time. Formalising this , we have the following: Observe that if $\delta=\text{dist}(z_0,\mathbb{S^1})$, and $|z-z_0|=h<\dfrac{\delta}{2}$, then $$|\zeta-z|=|(\zeta-z_0)-(z-z_0)|\geq|\zeta-z_0|-|z-z_0|>\dfrac{\delta}{2}$$
Therefore if $z\to z_0$, then $\dfrac{1}{|\zeta-z||\zeta -z_0|}>\dfrac{\delta^2}{2}$
Now you need only note that $$|\dfrac{F(z)-F(z_0)}{z-z_0}|\leq\dfrac{1}{2\pi}\int_{\mathbb{S^1}}\dfrac{|f(\zeta)|}{|\zeta-z||\zeta -z_0|}<\text{sup}_{z\in\mathbb{S^1}}|f(z)|\dfrac{2}{\delta^2}$$
Conclude!