 Solution

Assume the Continuum hypothesis and enumerate $\mathbb{R}=\{r_{\alpha }:\alpha <{\omega }_1\}$. DefineEvery 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