Need help with proving/understanding uniform convergence of derivatives of analytic function. (Proof Verification) 
I have proved the first part its the second part I'm having problems with.
Sketch of proof of second part:
Let $C$ be a compact set. It can be covered by finite $B_r(C_i)$ where $\forall i$ $\bar B_r(C_i)\subset U$. 
I want to show that $F_n'$ converges uniformly on each $B_r(C_i)$.
$\forall \epsilon >0$ there exists $N$ s.t for all $n\geq N$ and for all $z\in \bar B_r(C_i)$ $|f_n(z)-f(z)|<\epsilon$ (Uniform Convergence)
Hence, 
$\forall \epsilon >0$ there exists $N$ s.t for all $n\geq N$ and for all $l\in B_r(C_i)$
$|f_n'(l)-f'(l)|=|\frac{1}{2\pi i}\int_{C[c_i,r]}\frac{f_n(w)-f(w)}{(w-l)^2}dw|\leq\frac{r\epsilon}{(r-|l-c_i|)^2}$
However, i can seem to proceed from here due to the l term in the bound. Any ideas?
 A: First, I will prove that $(f_n')_{n\in\mathbb N}$ converges uniformly to $f'$ on every closed disk $\overline{D(z_0,\varepsilon)}\subset U$. Take $\varepsilon'\in(\varepsilon,+\infty)$ such that $\overline{D(z_0,\varepsilon')}\subset U$ and consider that loop $\gamma\colon[0,2\pi]\longrightarrow U$ defined by $\gamma(t)=z_0+\varepsilon'e^{it}$. Then\begin{align}\left(\forall z\in\overline{D(z_0,\varepsilon)}\right):\left\lvert f'(z)-f_n'(z)\right\rvert&=\frac1{2\pi}\left\lvert\int_\gamma\frac{f(u)-f_n(u)}{(u-z)^2}\,\mathrm du\right\rvert\\&\leqslant\varepsilon'\frac{\sup_{\lvert z'-z\rvert=\varepsilon'}\bigl\lvert f(z')-f_n(z')\bigr\rvert}{(\varepsilon'-\varepsilon)^2},\end{align}since the length of the loop $\gamma$ is $2\pi\varepsilon'$. Now, it follows from the fact that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$ on $\{z\in\mathbb{C}\,|\,\lvert z-z_0\rvert=\varepsilon'\}$ that $(f_n')_{n\in\mathbb N}$ converges uniformly to $f'$ on $\overline{D(z_0,\varepsilon)}$.
Now, to deduce that we have uniform convergence on every compact subset $K$ of $U$, we use the usual compactness argument: for each $z\in K$, there is some $\varepsilon_z>0$ such that $\overline{D(z,\varepsilon_z)}\subset K$. Since $K$ is compact, there are elements $z_1,\ldots,z_n\in  K$ such that $K\subset\bigcup_{j=1}^nD(z_j,\varepsilon_{z_j})$. Therefore…
