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 $D$.