Is every element of $\mathbb{R}$ a member of $\mathbb{Q}$ adjoined with finitely many members of its transcendence basis?

Question

Recently I've been interested in creating somewhat non-constructive solutions to problems using the concept of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, which exists assuming the Axiom of Choice but I only know some basic Field Theory. As part of my increasing understanding, I ask:

Let $W$ be the transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$. Is it true that $$\mathbb{R} = \bigcup_{w\subset W, \;w \text{ finite}}\mathbb{Q}(w)$$? What if we replace "finite" with "countable"?

Answer 1

Perhaps I am missing something, but, quoting for instance from this MSE post:

a set $T$ of elements of an extension field $k/F$ is a transcendence basis if

1. for all $n$, and distinct $t_{1}, \dots, t_{n} \in T$, there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(t_1,\dots,t_n)=0$;
2. $k$ is algebraic over $F(T)$.

So an element like $\sqrt{2}$ will not be in any of your $\mathbb{Q}(w)$.

Answer 2

Edit. This answer is incorrect. I read "transcendence basis" as "vector space basis". I think @AndreasCaranti 's answer is correct. I will leave mine up so no one else makes the same mistake.

---

Yes, since every element of $\mathbb{R}$ is a finite $\mathbb{Q}$-linear combination of basis elements. That means it's in the union of the corresponding extensions.