Proving that a sequence of functions converge uniformly

The convergence is not uniform on $(-1,1)$, but it is uniform on closed subsets thereof. We first examine the convergence on the closed subinterval $[-r,r]$ for $0<r<1$. After that, we show that the convergence fails to be uniform on $(-1,1)$

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Let $\epsilon>0$ be given. Then, we have for $x\in [-r,r]$ for $0<r<1$

$$\begin{align} \left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\ &\le \frac{r^{n+1}}{1-r}\\\\ &<\epsilon \end{align}$$

whenever $n>\frac{\log(1-r)+\log(\epsilon)}{\log(r)}-1$

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Take $\epsilon=1/2$. Then, for all $N$ we take $x=(1/2)^{1/(n+1)}$ for $0<r<1$ and any $n>N$. Then, we find that

$$\begin{align} \left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\ &= \frac{1/2}{1-(1/2)^{1/n+1}}\\\\ &\ge 1/2\\\\ &=\epsilon \end{align}$$

which negates the uniform convergence on the open interval $(-1,1)$

I think you are on the right way, note that:

$$ \left|S_n(x)- \frac{x}{1-x} \right|=\left| -\frac{x^{n+1}}{1-x} \right| \ge | x^{n+1} | > 0 $$ for every $x \in (0,1) $ and every $n \in \mathbb{N}$. Taking least upper bounds basically gives the solution.

So, you have $$ \left\lvert S_n(x) - S(x)\right\rvert = \left\lvert \frac{1-x^n}{1-x} - \frac{1}{1-x}\right\rvert= \left\lvert \frac{x^n}{1-x}\right\rvert $$ for all $x\in(-1,1)$.

Now, consider $x_n \stackrel{\rm def}{=} 1-\frac{1}{n}\in(-1,1)$. We have $$ \sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert \geq \left\lvert S_n(x_n) - S(x_n)\right\rvert =\left\lvert \frac{x_n^n}{1-x_n}\right\rvert = \frac{(1-\frac{1}{n})^n}{\frac{1}{n}} = n \left(1-\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{}\infty $$ so $\sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert\not\xrightarrow[n\to\infty]{}0$, meaning there is no uniform convergence on $(-1,1)$.