Family of univalent functions closed under normal convergence? I am encountering a question in complex analysis that is giving me somewhat of a hard time. Here it is:  

Say that $\mathcal{S}$ is a family of univalent functions $f(z)$ defined on the open unit disk $\{|z| <1 \}$ that satisfy $f(0)=0$ and $f'(0)=1$. We are asked to show that $\mathcal{S}$ is closed under normal convergence, i.e., if a sequence in $\mathcal{S}$ converges normally to $f(z)$, then $f \in \mathcal{S}$. 

Now by definition I know that a sequence of functions $f_n \rightarrow f$ normally on a domain $D$ if $f_n \rightarrow f$ uniformly on a compact subset of  $D$. Also, a function is univalent on a domain $D$ if it is one-to-one and analytic on $D$. So what approach should I take in order to prove this question? Any helpful input would be appreciated.
 A: Suppose $\{f_n\}$ is a sequence in $\mathcal{S}$ such that $f_n$ converges normally to $f$. Denote $D=\{z: |z|<1\}$, the open unit disk. 
Let $\gamma$ be any closed piecewise $C^1$ curve in $D$, which is a compact subset in $D$. So $f_n$ converges to $f$ uniformly on $\gamma$. This implies that 
$$\int_\gamma f=\int_\gamma\lim_{n\rightarrow\infty} f_n=\lim_{n\rightarrow\infty}\int_\gamma f_n=0,$$
where the last equality follows from Cauchy integral theorem that $\displaystyle\int_\gamma f_n=0$ for all $n$. Therefore, by Morera's theorem, $f$ is holomorphic. 
Next, we have $f(0)=0$, because 
$f(0)=\displaystyle\lim_{n\rightarrow\infty}f_n(0)=0$ since $f_n(0)=0$ for all $n$. On the other hand, since $\{z:|z|=r\}$ is a compact set in $D$ when $r>0$ is sufficiently small, $f_n$ converges to $f$ uniformly on $\{z:|z|=r\}$. This implies that 
$$f'(0)=\int_{\{|z|=r\}}\frac{f(z)}{z^2}=\int_{\{|z|=r\}}\lim_{n\rightarrow\infty}\frac{f_n(z)}{z^2}=\lim_{n\rightarrow\infty}\int_{\{|z|=r\}}\frac{f_n(z)}{z^2}=\lim_{n\rightarrow\infty}f_n'(0)=1$$
where the last two equalities follows from Cauchy's integral formula and the assumption that $f'_n(0)=1$ for all $n$. 
Similarly, using $\displaystyle f_n^{(m)}(z_0)=\int_{\{|z-z_0|=r\}}\frac{f(z)}{(z-z_0)^{m+1}}$ and $f^{(m)}(z)=\displaystyle\int_{\{|z-z_0|=r\}}\frac{f(z)}{(z-z_0)^{m+1}}$ for sufficiently small $r>0$, we can show that $f^{(m)}_n$ converges to $f^{(m)}$ on any compact subset in $D$, for any $m\in\mathbb{N}$. 
Since $f'(0)=1$, $f'$ is not identically zero in $D$, and the zeros of $f'$ must be isolated. Let $\gamma$ be a closed curve which does not pass through any zeros of $f$. Note also that $f_n'$ does not have any zero because $f_n$ is univalent by assumption. Then we have
$$\frac{1}{2\pi i}\int_\gamma\frac{f''}{f'}=\frac{1}{2\pi i}\int_\gamma\lim_{n\rightarrow\infty}\frac{f_n''}{f_n'}=\lim_{n\rightarrow\infty}\frac{1}{2\pi i}\int_\gamma\frac{f_n''}{f_n'}=0$$
where the last equality follows from the fact that $f_n'$ does not have any zero. Note also that $\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f''}{f'}$ is the number of zeros of $f'$ in region bounded by $\gamma$. Hence, $f'$ does not have any zero in $D$ since $\gamma$ is aribtrary, which implies that $f$ is univalent in $D$. 
