This is a theorem with a direct proof by construction.
The statement simply says "there exists a rearrangement". So the book assembles one. And it does not mean there cannot exist any other construction! The proof states one case, namely expression (25) and shows it works. That's it.
The structure of the proof is basically the following.
First, we put up Expression (25). Then we show that all its constituents indeed exist and make sense, namely $p_n, q_n, P_{m_k}, Q_{m_k}, \beta_n, \alpha_n$.
A picture is worth a thousand words.
Curves in blue and in green are sequences $\beta_n$ and $\alpha_n$, they converge to $\beta$ and $\alpha$ respectively.
Curve in red is our Expression (25). The way we painted this picture shows a few things about Expression (25).
The first terms of it are positive, so the reason why $\beta_1 >0$, and the curve first goes up. Again, in our construction we first put Expression (25) as a basis, starting point of the proof. The first terms of it are positive, so $\beta_1$ must be positive. We tweak betas and other components to make them tie in with Expression (25).
Next, $x_n > \beta_n$ and $y_n < \alpha_n$, so the red curve of Expression (25) sticks out a bit "outside" sequences $\beta_n$ and $\alpha_n$. By one last positive term and negative term $P_{m_n}$ and $Q_{k_n}$.
Differences between Expression (25) and these sequences (marked X) become smaller because $P_n \to 0$ and $Q_n \to 0$. It is precisely for this reason it becomes finally "clear" that Expression (25) cannot converge to any number greater than $\beta$ or smaller than $\alpha$.
Can we have an alternative construction? Why not!
Just change Expression (25), and tweak all other elements accordingly, e.g. make Expression (25) "expand" inside the "space" between $\beta$ and $\alpha$. Make the first terms negative numbers, the curve would first go down in this case. The differences (marked Y) would also gradually become smaller. And the whole thing would work as well.
You might ask why the theorem is stated only for the real numbers? That's because all inequalities do not make sense for the complex numbers, which do not have '<' defined.
For $i=1,2$ let $\varphi_i:\Bbb N\to M_i$ be a bijection, and let
$$\varphi:\Bbb N:M_1\cup M_2:n\mapsto\begin{cases}
\varphi_1\left(\frac{n}2\right),&\text{if }n\text{ is even}\\
\varphi_2\left(\frac{n-1}2\right),&\text{if }n\text{ is odd.}
\end{cases}$$
(Note that for me $\Bbb N$ includes $0$.) It’s not hard to verify that $\varphi$ is a bijection such that
$$\langle\varphi(0),\varphi(1),\varphi(2),\varphi(3),\ldots\rangle=\langle\varphi_1(0),\varphi_2(0),\varphi_1(1),\varphi_2(1),\ldots\rangle\,,$$
so that
$$\sum_{k=0}^{2n+1}a_{\varphi(k)}=\sum_{k=0}^na_{\varphi_1(k)}+\sum_{k=0}^na_{\varphi_2(k)}$$
for each $n\in\Bbb N$.
Let
$$\ell=\sum_{k\in M_1\cup M_2}a_k=\sum_{k\in\Bbb N}a_{\varphi(k)}$$
and
$$\ell_i=\sum_{k\in M_i}a_k=\sum_{k\in\Bbb N}a_{\varphi_i(k)}$$
for $i=1,2$; we want to show that $\ell=\ell_1+\ell_2$.
Let $\epsilon>0$; there is an $n_\epsilon\in\Bbb N$ such that
$$\left|\ell_i-\sum_{k=0}^{n_\epsilon}a_{\varphi_i(k)}\right|<\frac{\epsilon}2$$
for $i=1,2$. Then
$$\begin{align*}
\left|(\ell_1+\ell_2)-\sum_{k=0}^{2n+1}a_{\varphi(k)}\right|&=\left|(\ell_1+\ell_2)-\left(\sum_{k=0}^na_{\varphi_1(k)}+\sum_{k=0}^na_{\varphi_2(k)}\right)\right|\\
&\le\left|\ell_1-\sum_{k=0}^na_{\varphi_1(k)}\right|+\left|\ell_2-\sum_{k=0}^na_{varphi_2(k)}\right|\\
&<\frac{\epsilon}2+\frac{\epsilon}2\\
&=\epsilon\,,
\end{align*}$$
so
$$\ell=\lim_{n\to\infty}\sum_{k=0}^{2n+1}a_{\varphi(k)}=\ell_1+\ell_2\,.$$
Best Answer
There is a rigorous and (to me) surprisingly simple proof that closely follows the argument in the book, but it is hard to be sure that it is what the author had in mind, because the quoted extract does not define the meaning of $\sum_{i \in I}a_i$ when $I$ is an infinite subset of $\mathbb{N} = \{1, 2, 3, \ldots\}$, and it would be interesting to know if a definition was given earlier in the book.
Such a sum may be defined in several ways (some of them quite elaborate), but all give the same value. I choose a definition that is simple and convenient for this question. For all $n \in \mathbb{N},$ define $[n] = \{1, 2, \ldots, n\}.$ Define $\sum_{i \in I}a_i$ to be the limit of the sequence of finite partial sums $\sum_{i \in I \cap [n]}a_i,$ as $n$ tends to infinity, if this limit exists. We could restrict $n$ to take values in $I,$ but it is convenient to allow any $n \in \mathbb{N},$ because then the result remains true even when $I$ is finite. (The sequence of partial sums is then eventually constant.) This reduces the need to consider special cases. It may be reassuring to note that when $I$ is infinite, the definition is equivalent to $\sum_{i \in I}a_i = \sum_{r=1}^\infty a_{h(r)},$ where $h \colon \mathbb{N} \to \mathbb{N}$ is the unique strictly increasing function such that $h(\mathbb{N}) = I.$ Thus $\sum_{i \in I}a_i$ is the sum of a 'subseries' of $\sum_{i=1}^\infty a_i$ (but I don't think this is a technical term, unlike 'subsequence').
Let $I, J$ be disjoint subsets of $\mathbb{N}$ such that $\sum_{i \in I}a_i, \sum_{i \in J}a_i$ exist. Then the following limit also exists: \begin{align*} \sum_{i \in I \cup J}a_i & = \lim_{n \to \infty} \sum_{i \in (I \cup J) \cap [n]}a_i \\ & = \lim_{n \to \infty} \sum_{i \in (I \cap [n]) \cup (J \cap [n])}a_i \\ & = \lim_{n \to \infty} \left( \sum_{i \in I \cap [n]}a_i + \sum_{i \in J \cap [n]}a_i \right) \\ & = \left( \lim_{n \to \infty} \sum_{i \in I \cap [n]}a_i \right) + \left( \lim_{n \to \infty} \sum_{i \in J \cap [n]}a_i \right) \\ & = \sum_{i \in I}a_i + \sum_{i \in J}a_i. \end{align*} It follows by induction on $m \in \mathbb{N}$ that if $I_1, I_2, \ldots, I_m$ is a finite sequence of $m$ pairwise disjoint subsets of $\mathbb{N}$ such that $\sum_{i \in I_k}a_i$ exists for $k = 1, 2, \ldots, m,$ then \begin{equation}\label{4189981:eq:1}\tag{$1$} \sum_{i \in I_1 \cup I_2 \cup \cdots \cup I_m} \!\! a_i = \sum_{i \in I_1}a_i + \sum_{i \in I_2}a_i + \cdots + \sum_{i \in I_m}a_i. \end{equation}
Now consider the case of an infinite sequence $(I_k)_{k \in \mathbb{N}}$ of pairwise disjoint subsets of $\mathbb{N}.$ We begin to make use of the hypothesis that the series $\sum_{i=1}^\infty a_i$ is absolutely convergent. Because of this hypothesis, we do not need to postulate additionally that $\sum_{i \in I_k}a_i$ exists for all $k \in \mathbb{N}.$ In fact, it is clear that for any subset $I \subseteq \mathbb{N},$ and any $n \in \mathbb{N},$ $$ \sum_{i \in I \cap [n]}|a_i| \leqslant \sum_{i=1}^n|a_i|, $$ therefore if $I$ is infinite, the 'subseries' $\sum_{r=1}^\infty a_{h(r)}$ (where the function $h \colon \mathbb{N} \to \mathbb{N}$ is as defined above) is absolutely convergent, whence $\sum_{i \in I}a_i$ exists.
(We could have avoided some slight awkwardness of expression here by defining a concept of 'absolute convergence' especially for expressions of the form $\sum_{i \in I}a_i,$ so that the function $h$ need not be mentioned, and we need only observe that $\sum_{i \in I}|a_i|$ exists, therefore $\sum_{i \in I}a_i$ exists. Although that would have been easy to do, it hardly seems worth even the small amount of trouble.)
It would be interesting to know if the author of the textbook quoted in the question was taking all of that argument - or something similar - for granted, as being 'obvious', or whether it was stated explicitly earlier in the book.
Let $J = \bigcup_{k=1}^\infty I_k.$ (The question concerns the case $J = \mathbb{N},$ but it is not much harder to consider the general case.) For all $n \in \mathbb{N},$ $$ J \cap [n] = \bigcup_{k=1}^\infty (I_k \cap [n]). $$ The finite set $J \cap [n]$ can only be a disjoint union of finitely many distinct, non-empty sets, therefore there exists $m \in \mathbb{N}$ such that $$ J \cap [n] = \bigcup_{k=1}^m (I_k \cap [n]) = (I_1 \cup I_2 \cup \cdots \cup I_m) \cap [n]. $$ Define $$ K = \! \bigcup_{k=m+1}^\infty I_k \subseteq J \setminus [n]. $$ Then $$ J = I_1 \cup I_2 \cup \cdots \cup I_m \cup K, $$ and this is a union of pairwise disjoint sets, therefore, by \eqref{4189981:eq:1}, $$ \sum_{i \in J}a_i = \sum_{i \in I_1}a_i + \sum_{i \in I_2}a_i + \cdots + \sum_{i \in I_m}a_i + \sum_{i \in K}a_i, $$ whence \begin{equation}\label{4189981:eq:2}\tag{$2$} \left\lvert \sum_{i \in J}a_i - \sum_{i \in I_1}a_i - \sum_{i \in I_2}a_i + \cdots - \sum_{i \in I_m}a_i \right\rvert = \left\lvert \sum_{i \in K}a_i \right\rvert \leqslant \sum_{i \in J \setminus [n]} |a_i|. \end{equation}
We pause briefly, to justify the last step in \eqref{4189981:eq:2}. (Hopefully the rest is clear!)
Let $E, F$ be subsets of $\mathbb{N}$ (finite or infinite) such that $E \subseteq F$ and $\sum_{i \in F}|a_i|$ exists. (Here, we temporarily drop the hypothesis that $\sum_{i=1}^\infty a_i$ is absolutely convergent: for now, it may be any series at all.) Then for all $n \in \mathbb{N},$ $$ \sum_{i \in E \cap [n]}|a_i| \leqslant \sum_{i \in F \cap [n]}|a_i|. $$ By hypothesis, the partial sums on the right are bounded above. Therefore $\sum_{i \in E}|a_i|$ exists. Again using the rather awkward identification of expressions like $\sum_{i \in E}a_i$ with sums $\sum_{r=1}^\infty a_{h(r)}$ of 'subseries' of $\sum_{i=1}^\infty a_i,$ we conclude that $\sum_{i \in E}a_i$ also exists (is 'absolutely convergent' - perhaps after all it would have been worth the effort of making this terminology usable literally). This is in the case where $E$ (therefore also $F$) is infinite; if $E$ is finite, then of course there is nothing to worry about. Finally, we have for all $n \in \mathbb{N},$ $$ \bigg\lvert \sum_{i \in E \cap [n]}a_i \bigg\rvert \leqslant \sum_{i \in E \cap [n]}|a_i| \leqslant \sum_{i \in F \cap [n]}|a_i|, $$ whence, taking limits as $n \to \infty,$ \begin{equation}\label{4189981:eq:3}\tag{$3$} \bigg\lvert \sum_{i \in E}a_i \bigg\rvert \leqslant \sum_{i \in F}|a_i|. \end{equation}
Using \eqref{4189981:eq:3} with $E = K$ and $F = J \setminus [n],$ we have fully justified \eqref{4189981:eq:2}.
Because $\sum_{i=1}^\infty a_i$ is absolutely convergent (actually it would have been enough to postulate that $\sum_{i \in J}|a_i|$ exists, but this is a minor point, and I shall carry on with the more extravagant hypothesis), for every $\epsilon > 0,$ there exists $n \in \mathbb{N}$ such that $\sum_{i=n+1}^\infty|a_i| < \epsilon.$ Using \eqref{4189981:eq:3} again (this awkwardness could have been avoided, either by using the weaker hypothesis just mentioned, or else by treating only the case $J = \mathbb{N}$), it follows that $\sum_{i \in J \setminus [n]}|a_i| < \epsilon.$ By \eqref{4189981:eq:2}, therefore, $$ \left\lvert \sum_{i \in J}a_i - \sum_{i \in I_1}a_i - \sum_{i \in I_2}a_i + \cdots - \sum_{i \in I_m}a_i \right\rvert < \epsilon. $$ Since $\epsilon$ is arbitrary, we have proved $$ \sum_{i \in J}a_i = \sum_{k=1}^\infty\sum_{i \in I_k}a_i. $$