= Solution
The <Freiling theorem> gives
$$
\boxed{A_{<\omega_1}(\mathbb R)\ \Longleftrightarrow\ \neg\mathrm{CH}}.
$$
First assume the <Continuum hypothesis> and fix a <well-order> $(r_\alpha)_{\alpha<\omega_1}$ of the <real numbers>. Define $f(r_\alpha)=\{r_\beta:\beta\leq\alpha\}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the <Freiling axiom of symmetry>.
Conversely, assume $|\mathbb R|>\aleph_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq\mathbb R$ of <cardinality> $\aleph_1$. By <infinite cardinal arithmetic>,
$$
\left|X\cup\bigcup_{x\in X}f(x)\right|\leq\aleph_1<|\mathbb R|.
$$
Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the <Freiling axiom of symmetry> and completes both implications.
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