This is Theorem 8.2 from baby Rudin, where it is assumed that series given has radius of convergence $1$. :
I can't understand last couple of lines in proof fully. In last line we have to derive $$(1-x)\sum_{n=0}^{N}|s_n-s||x|^n<\frac{\varepsilon}{2}$$ for some $\delta>0$ with $x>1-\delta$, and $$\left|(1-x)\sum_{n=N+1}^{\infty}(s_n-s)x^n\right|<\frac{\varepsilon}{2}.$$ Here, second inequality, I think, holds because, if $\sum_{i=1}^\infty a_i$ is convergent series, then $\sum_{i=n}^\infty a_n\to0$ as $n\to0$. How to get first inequality?
Since $|x| < 1$ it follows that
$$\sum_{n=0}^{N}|s_n-s||x|^n < \sum_{n=0}^{N}|s_n-s| = A$$
where $A$ is a constant for fixed $N$ independent of $x$.
Thus, with $x < 1$, we have as $x \to 1-$,
$$0 \leqslant (1-x)\sum_{n=0}^{N}|s_n-s||x|^n < A(1-x) \to 0$$
You are on the right track for the second inequality since $|s_n - s| < \epsilon$ for all sufficiently large $n$, and for $0 < x < 1$
$$\sum_{n=N+1}^{\infty}|x|^n = \frac{x^{N+1}}{1-x}$$
Try to finish from here.