= Solution
Let $f$ be given and choose $Y\subseteq\mathbb R$ of <cardinality> $\aleph_1$. Each $f(y)$ is countable, so the <axiom of choice> and <infinite cardinal arithmetic> give
$$
\left|Y\cup\bigcup_{y\in Y}f(y)\right|\le\aleph_1.
$$
Since the continuum is larger, choose $x$ outside this union. Then $x\notin f(y)$ for every $y\in Y$. The countable <set> $f(x)$ cannot contain $Y$, so choose $y\in Y\setminus f(x)$. Thus
$$
\boxed{x\notin f(y),\qquad y\notin f(x).}
$$
The two points are distinct because $x\notin Y$. This proves the <countable-valued free-pair criterion> and the required direction of the <Freiling axiom of symmetry> without any measurability assumption on $f$.
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