Lipschitz type property for analytic function on unit disk

Question

Let $f$ be an analytic function on the open disk $b_1(0)$. Assume that $f'$ is continuous and bounded. Prove that there exists some $C > 0$ such that
$$
|f(z_1)-f(z_2)| \leq C|z_1-z_2| \forall z_1,z_2 \in b_1(0).
$$

I can see the following: if $D$ is any closed disk contained in $b_1(0)$ and $z_1, z_2 \in D$, then using the compactness of the closed disk and the boundedness of $f'$, then $\frac{f(z_1)-f(z_2)}{z_1-z_2}$ is bounded above. But I not sure how to extend this to the whole $b_1(0)$.

Here is a similar problem, but $f$ is assumed to be bounded here.

Answer 1

HINT:

1. If $\gamma$ is a curve on $b_1(0)$ that begins at $z_1$ and ends at $z_2$ then $\int_{\gamma} f'(z)\;dz = f(z_2)-f(z_1)$
2. $$\left|\int_\gamma g(z) \;dz\right|\; \leq \; \sup_{z\in [\gamma]}|g(z)| \cdot l(\gamma)$$ where $l(\gamma)$ is the length of the curve $\gamma$.

Can you put these two together to complete the solution? Notice that the continuity of $f'$ is not needed.