Let me repeat what I said in the comments. There is no ambiguity here. We have the standard notation for the field of fractions $K(x_1,x_2,\dots,x_n)$, and we have a definition of what $K(B)$ means for a set.
The question remains whether $K(x_1,x_2,\dots,x_n)=K(\{x_1,x_2,\dots,x_n\})$. The answer is "of course" and here are some of the details.
Let $K$ be a field, $R:=K[x_1,x_2,\dots, x_n]$ be the polynomial ring in $n$ variables.
Let $L$ denote the field of fractions of $R$, so that $L=\left\{\frac{f}{g}\mid f,g\in R, g\ne0\right\}$. This field is usually denoted by $K(x_1,x_2,\dots,x_n)$.
Now let $B$ be the set $\{x_1,x_2,\dots,x_n\}$. Then consider the field $K(B)$ defined as the smallest subfield of $L$ containing $K$ and the set $B$.
By definition we have that $K(B)\subseteq L$.
Now let $\frac{f}{g}\in L$. We have that $f,g\in K[x_1,x_2,\dots,x_n]$. We can write $f=\sum_{} a_{i_1,i_2,\dots,i_n} x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$ where the ceofficients are in $K$ and the sum is finite. Now $K(B)$ by definition contains all the $x_i$ and so (being closed under multiplication) contains all the monomials $x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$. $K(B)$ also by definition contains all the coefficients as these are in $K$. Hence, as $K(B)$ is a field it must contain $f$. The same is true for $g$. But $K(B)$ is a field so it must contain also $\frac{f}{g}$. That is $L\subseteq K(B)$.
We therefore have that $L=K(B)$.
That is, using the usual notation for the field of fractions,
$$
K(x_1,x_2,\dots, x_n)=K(\{x_1,x_2,\dots, x_n\})
$$
and we are done.
Not only are the real numbers the smallest complete ordered field containing $\mathbb{Q}$, they are in fact the only complete ordered field. The proof is surprisingly straightforward: you show that any two such have to have the same prime subfield, and then you lift the isomorphism between their prime subfields to one between the fields themselves. You can find the full details of the proof here.
As mentioned in a comment, the issue with "Analysis in the Computable Number Field" is that they're using a different notion of completeness than "mainstream" non-constructive analysis. If I were to describe a sequence that doesn't converge to an element of Aberth's field, Aberth would object that in his axioms the sequence does not in fact converge. And he would be right, but he's using a different set of axioms than the majority of mathematicians.
Best Answer
Define a ring homomorphism $f\colon \mathbb{Q}[x]\to\mathbb{R}$ by $$f(p)=p(\pi)$$ so that (for example) $$f(\tfrac{1}{3})=\tfrac{1}{3},\quad f(x)=\pi,\quad f(2x^2+5)=2\pi^2+5,$$ and so on. The image of a ring homomorphism is a subring of the codomain; in the case of this particular ring homomorphism $f$, the image is given the name $\mathbb{Q}[\pi]$. It is the "smallest" subring of $\mathbb{R}$ that contains $\mathbb{Q}$ and $\pi$.
What is the kernel of this homomorphism $f$? That is, what polynomials $p\in\mathbb{Q}[x]$ have $\pi$ as a root? The answer is none (other than the obvious $p=0$). This is what it means for $\pi$ to be transcendental (in 1882, Lindemann proved that $\pi$ is transcendental). The first isomorphism theorem for rings now tells us that $$\mathbb{Q}[x]/(\ker f)\cong \mathbb{Q}[\pi]$$ but since the kernel of $f$ is trivial, this statement just says that $\mathbb{Q}[x]\cong \mathbb{Q}[\pi]$. In other words, we see that $f$ is a ring isomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[\pi]$.
Now see if you can prove the following general fact: if two integral domains $D_1$ and $D_2$ are isomorphic, then their respective fields of fractions $\mathrm{Frac}(D_1)$ and $\mathrm{Frac}(D_2)$ are also isomorphic.