= Solution
Assume the <Continuum hypothesis> and enumerate $\mathbb R=\{r_\alpha:\alpha<\omega_1\}$. Define
$$
f(r_\alpha)=\{r_\beta:\beta\le\alpha\}.
$$
Every value is countable. For any two indices, say $\alpha\le\beta$, one has $r_\alpha\in f(r_\beta)$, so the two required nonmembership conditions cannot both hold. Including the endpoint in each initial segment also rules out taking the two points equal. Therefore the free-pair assertion implies the negation of <Continuum hypothesis>. <Cantor theorem> and choice already give $2^{\aleph_0}\ge\aleph_1$, so
$$
\boxed{A_{<\aleph_1}(\mathbb R)\ \Longleftrightarrow\ 2^{\aleph_0}>\aleph_1\quad\text{in ZFC}.}
$$
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