Here is a more analytical proof.
First we prove that the function is continuous on $\mathbb{C}\setminus[0,1]$. Since $\phi$ is contiuous and $[0,1]$ is compact, $|\phi|$ attains some maximum, say $M$. Also $\mathbb{C}\setminus[0,1]$ is open. Fix some $z_0$ such that $D(z_0, 2r) \subseteq \mathbb{C}\setminus[0,1]$. Then $|f(z)-f(z_0)|=\left|\int_0^1\frac{\phi(t)}{t-z} - \frac{\phi(t)}{t-z_0}\ dt\right|$
$=\left|\int_0^1\phi(t)\cdot\frac{z-z_0}{(t-z)(t-z_0)}\ dt\right|$ from the $LM$ estimate, we then have as $\epsilon \rightarrow 0$ $\le |z-z_0|\cdot \max_{t\in[0,1]}\left|\frac{\phi(t)}{(t-z)(t-z_0)}\right|$ Taking $|z-z_0|<\epsilon we have $\le\frac{M}{2r^2}\epsilon \rightarrow 0$ With continuity at hand, consider then the differential quotient $\frac{f(z)-f(z_0)}{z-z_0}=\int_0^1\frac{\phi(t)}{(t-z)(t-z_0)}\ dt$ now the function $\varphi(t) \equiv \frac{\phi(t)}{t-z_0}$ is also continuous. So that as $z\rightarrow z_0$ $\int_0^1\frac{\varphi(t)}{(t-z)}\ dt\rightarrow \int_0^1\frac{\varphi(t)}{(t-z_0)}\ dt$ which shows that $f'(z) = \int_0^1\frac{\phi(t)}{(t-z_0)^2}\ dt$