A real number is algebraically constructible exactly when it lies in a finite tower of real quadratic adjunctions

Let $T$ be the union of all terminal fields of finite towers obtained from $\mathbb{Q}$ by adjoining positive square roots, omitting any adjunction whose square root is already present. The empty tower puts $\mathbb{Q}$ in $T$.

Conversely, induction along any such tower shows $K_i\subseteq \mathcal{C}$: the base $\mathbb{Q}$ lies in $\mathcal{C}$, and square-root closure puts each generator and hence its generated field in $\mathcal{C}$. Thus $T\subseteq \mathcal{C}$.

If $x,y\in T$, append the generators of their two towers to one another. Any redundant adjunction is omitted; every remaining step has degree $2$, because its generator $\sqrt{a}$ is a root of $t^2-a$ over the base, so by [L4] its minimal polynomial divides $t^2-a$ and has degree at most $2$, while degree $1$ would put $\sqrt{a}$ in the base, which the omission of redundant adjunctions excludes; [L2] then gives $[K_{i-1}(\sqrt{a}):K_{i-1}]=2$. Thus $x-y$, $xy$, and $x^{-1}$ for $x\mathrm{\ne }0$ lie in a common terminal field, so $T$ is a subfield.

If $0<a\in T$, append $\sqrt{a}$ to a tower containing $a$, or do nothing if it is already present. Hence $T$ is closed under positive square roots. By minimality in [L1], $\mathcal{C}\subseteq T$.

Therefore $T=\mathcal{C}$. Membership in $T$ is exactly the existence of a displayed finite quadratic tower, including the zero-length case.