About a proof of Abel's theorem by Kazuo Matsuzaka in "A Course in Analysis vol.2" (in Japanese). I cannot derive an inequality in his proof. I am reading "A Course in Analysis vol.2" (in Japanese) by Kazuo Matsuzaka.
There is the following theorem (Abel's theorem) in this book:

Let $\sum a_n x^n$ be a power series whose radius of convergence is $1$.
Let $f(x) := \sum_{n=0}^{\infty} a_n x^n$ for $|x|<1$.
Suppose that $\sum_{n=0}^\infty a_n$ converges.
Then, $\lim_{x\to 1} f(x)=\sum_{n=0}^\infty a_n$ holds.

And the author's proof of this theorem is here:

Let $s_n := a_0+a_1+\dots+a_n$.
Let $s:=\lim_{n\to\infty}s_n=\sum_{n=0}^\infty a_n$.
Let $|x|<1$.
$$a_0 + a_1 x+a_2 x^2+\dots+a_n x^n=s_0+(s_1-s_0)x+(s_2-s_1)x^2+\dots+(s_n-s_{n-1})x^n=(1-x)(s_0+s_1 x+\dots+s_{n-1} x^{n-1})+s_n x^n.$$
So, $f(x)=(1-x)\sum_{n=0}^\infty s_n x^n.$
Since $1=(1-x)\sum_{n=0}^\infty x^n$, $s=(1-x)\sum_{n=0}^\infty s x^n$.
$f(x)-s=(1-x)\sum_{n=0}^\infty(s_n-s)x^n$.
Let $\epsilon$ be any positive real number.
Then there exists an positive integer $N$ such that if $n>N$, then $|s_n-s|<\frac{\epsilon}{2}$.
If $|x|<1$, then $$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s||x|^n+\frac{\epsilon}{2}|(1-x)\sum_{n=N+1}^\infty x^n|<(1-x)\sum_{n=0}^{N}|s_n-s|+\frac{\epsilon}{2}.$$
Since $\sum_{n=0}^{N}|s_n-s|$ is a constant real number, if $0<1-x<\delta$ for sufficiently small positive real number $\delta$, then $$(1-x)\sum_{n=0}^N|s_n-s|<\frac{\epsilon}{2}$$ holds.
So, if $0<1-x<\delta$, then $$|f(x)-s|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
This means $$\lim_{x\to 1}f(x)=s.$$

I cannot derive the following inequality:

If $|x|<1$, then
$$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s||x|^n+\frac{\epsilon}{2}|(1-x)\sum_{n=N+1}^\infty x^n|$$ holds.

By the way, we are interested in $x$ which is very close to $1$, so we can assume that $x$ is positive.

If $0<x<1$, then
$$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)|\sum_{n=N+1}^\infty (s_n-s)x^n|\\\leq (1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)\sum_{n=N+1}^\infty |s_n-s|x^n\\\leq (1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)\sum_{n=N+1}^\infty \frac{\epsilon}{2}x^n\\= (1-x)\sum_{n=0}^N|s_n-s|x^n+\frac{\epsilon}{2}(1-x)\sum_{n=N+1}^\infty x^n$$ holds.

 A: For $\lvert x \rvert < 1$ we have
$$\frac{f(x)-s}{1-x}=\sum_{n=0}^\infty(s_n-s)x^n .$$
I have chosen this form to remove the factor $(1-x)$ from further transformation of the RHS. But you can also do everything with the original equation. We get
$$\left\lvert \frac{f(x)-s}{1-x} \right\rvert = \left\lvert\sum_{n=0}^\infty(s_n-s)x^n \right\rvert \le \left\lvert\sum_{n=0}^N(s_n-s)x^n \right\rvert + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \\ \le  \sum_{n=0}^N\lvert s_n-s\rvert \lvert x \rvert^n  + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le  \sum_{n=0}^N\lvert s_n-s\rvert  + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert .$$
But $\sum_{n=N+1}^\infty(s_n-s)x^n$ is absolutely convergent because $\lvert (s_n-s)x^n \rvert \le \frac{\epsilon}{2}\lvert x \rvert^n$ and we have
$$\left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \sum_{n=N+1}^\infty\frac{\epsilon}{2}\lvert x \rvert^n = \frac{\epsilon}{2}\sum_{n=N+1}^\infty\lvert x \rvert^n \le \frac{\epsilon}{2}\sum_{n=0}^\infty\lvert x \rvert^n = \frac{\epsilon}{2} \frac{1}{1 - \lvert x \rvert} .$$
Therefore for $0 \le x < 1$ we get
$$\lvert f(x)-s \rvert \le (1 - x)\rvert\sum_{n=0}^N\lvert s_n-s\rvert + \frac{\epsilon}{2} .$$
Note, however, that your text claims that for $\lvert x \rvert < 1$
$$\left\lvert(1-x)\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \frac{\epsilon}{2}\left\lvert (1-x)\sum_{n=N+1}^\infty x^n \right\rvert ,$$
i.e.
$$\left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \frac{\epsilon}{2}\left\lvert \sum_{n=N+1}^\infty x^n \right\rvert .$$
Here I have some doubts: If $-1 < x < 0$ and the sequence $(s_n-s)$ has alternating signs, the all terms in $\sum_{n=N+1}^\infty(s_n-s)x^n$ have the same sign whereas $\sum_{n=N+1}^\infty x^n$ is an alternating sequence. Anyway, the range $-1 < x < 0$ is irrelevant.
