 Solution

Start with $S_0=\mathbb{Q}$. Given countable $S_n$, let $S_{n+1}$ be the field generated by $S_n\cup \{\sqrt{|x|}:x\in S_n\}$. It is countable because its elements are values of rational expressions in finitely many members of a countable set. Then $S={\bigcup }_n{S}_n$ is countable and has (i)–(iii). Any other set with those properties contains every $S_n$ by induction, proving minimality and uniqueness.