In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$.
A crucial point in the trick is the existence of a transcendence basis of $\mathbb R$ over $\mathbb Q$.
So this necessitates the axiom of choice and I would be grateful to a savvy user for confirming (or infirming) my feeling that the existence of such an $A$ is indeed only possible under acceptance of the axiom of choice.