Start with $\mathbb{Q}$ and repeatedly adjoin real roots of irreducible polynomials of odd degree by transfinite recursion until it is no longer possible to do so. The resulting field $F\subset\mathbb{R}$ will have the property that every finitely generated subfield has odd degree over $\mathbb{Q}$, and so in particular $\sqrt{2}\not\in F$. Every odd degree polynomial over $F$ has a root, since it must have some irreducible factor of odd degree, and then that irreducible factor can only be linear since otherwise we would have adjoined a root of it in the construction of $F$.
(Actually, the transfinite recursion can be avoided, since by a suitable dovetailing construction you can arrange that every polynomial of odd degree gets handled within the first $\omega$ steps.)
Turning my comments into an answer:
When Hodges writes "an isomorphism $e\colon \langle\overline{a}\rangle_A\to B$", this is a typo. It should say either "an embedding $e\colon \langle\overline{a}\rangle_A\to B$" or "an isomorphism $e\colon \langle\overline{a}\rangle_A\to \langle\overline{b}\rangle_B$".
Indeed, since $\overline{a}$ generates $\langle \overline{a}\rangle_A$, if $\overline{b}$ fails to generate $B$, there can be no isomorphism $\langle\overline{a}\rangle_A\to B$ mapping $\overline{a}$ to $\overline{b}$.
To finish the proof, the key point is that $\overline{a}$ is a tuple from $A$ and $\overline{b}$ is a tuple from $B$ of the same length, and $\overline{a}$ and $\overline{b}$ satisfy the same literals, then there is an isomorphism $e\colon \langle\overline{a}\rangle_A\to \langle\overline{b}\rangle_B$ mapping $\overline{a}$ to $\overline{b}$ (and so $e$ can also be viewed as an embedding $\langle\overline{a}\rangle_A\to B$).
We define the isomorphism by $e(t^A(\overline{a})) = t^B(\overline{b})$ where $t(\overline{x})$ is an arbitrary term. This is well-defined, since $a'\in \langle \overline{a}\rangle_A$ if and only if there is some term $t(\overline{x})$ such that $a' = t^A(\overline{a})$, and if $t^A(\overline{a}) = s^A(\overline{a})$, then $t^B(\overline{b}) = s^B(\overline{b})$, since $\overline{a}$ and $\overline{b}$ satisfy the same literals. Similar reasoning shows that $t$ is a bijection, with inverse $e^{-1}(t^B(\overline{b})) = t^A(\overline{a})$.
For any $n$-ary function symbol $f$, we have $e(f^A(t_1^A(\overline{a}),\dots,t_n^A(\overline{a}))) = f^B(t_1^B(\overline{b}),\dots,t_n^B(\overline{b})) = f^B(e(t_1^A(\overline{a})),\dots,e(t_n^A(\overline{a})))$, so $e$ preserves $f$. And for any $n$-ary relation symbol $R$, we have $A\models R(t_1^A(\overline{a}),\dots,t_n^A(\overline{a}))$ if and only if $B\models R(t_1^B(\overline{b}),\dots,t_n^B(\overline{b}))$, since $\overline{a}$ and $\overline{b}$ satisfy the same literals. So $e$ is an isomorphism.
Best Answer
It is a fact that a field $R$ is real closed if and only if $R$ is not algebraically closed, but $R[\sqrt{-1}]$ is algebraically closed. (The implication from the definition you gave of real closed field to $R[\sqrt{-1}]$ being algebraically closed is the content of the "algebraic" proof of the fundamental theorem of algebra: once you verify that $\mathbb{R}$ is real closed by your definition, it follows from this characterization that $\mathbb{C}$ is algebraically closed.)
Now if $R$ is real closed, then $R[\sqrt{-1}]$ is the only proper algebraic extension of $R$, since any algebraic extension of $R$ embeds over $R$ in the algebraic closure $R[\sqrt{-1}]$, but there are no intermediate fields in this extension of degree $2$. Thus, if $R\subseteq R'$ are real closed fields, no proper algebraic extension of $R$ is contained in $R'$ (since $-1$ has no square root in $R'$), so every element of $R'$ which is algebraic over $R$ is already contained in $R$.
I'm not sure if this is the explanation Hodges had in mind, since I don't think he states or proves the characterization above in his book. But I do think it's a good way of thinking about this argument.