To clarify the problem, consider the field ${\Bbb R}$ as a field extension of ${\Bbb Q}$ using some sort of Hamel basis. Does there exist a field $F\subsetneq{\Bbb R}$ such that $F(\sqrt{2})={\Bbb R}$ (or $F(x)={\Bbb R}$ for any other irrational $x$)? If $(z_\alpha)_{\alpha<{\frak c}}$ is a basis for ${\Bbb R}$ as a ${\Bbb Q}$-vector space such that $z_0=\sqrt 2$, then $F={\Bbb Q}(z_1,z_2,\dots,z_\alpha,\dots)$ (where all $z_\alpha$ are adjoined for $0\ne \alpha<{\frak c}$) should be a field not containing $\sqrt2$, even though $F(\sqrt2)={\Bbb R}$. (I am using that $\sqrt{2}$ is algebraic of degree $2$ in this argument to prove that $\sqrt{2}\notin F$, although it is probably not necessary.)
My question relates to the process used to find this field. As you can see, I've used a nonconstructive proof using a Hamel basis (and the axiom of choice, implicitly), and this is not philosophically satisfactory for me. Is there an "explicit" proof of existence of such a field $F$, and do its elements have any nice characterization? Is $F$ uniquely determined by the properties $F\subsetneq{\Bbb R}$ and $F(\sqrt{2})={\Bbb R}$? What are the closure properties of $G:={\Bbb R}\setminus F$?
Edit: it seems the construction above doesn't work, and there are good reasons why the $F$ described above doesn't exist. I would like to patch my construction instead by specifying only that $F\subseteq{\Bbb R}$ and $\sqrt2\notin F$, and $F$ is maximal in the sense that there is no proper field extension $F(\alpha)$ satisfying the same properties. Surely this definition will work by Zorn's lemma, although I doubt that such an $F$ is unique. Am I right?