Axiom of Symmetry, aka Freiling's argument against CH

The point is that violations of the Axiom of Symmetry are fundamentally connected with non-measurable sets, and counterexample functions $f$ to AS cannot be nice measurable functions.

You have proved the one direction $CH\to \neg AS$, that if there is a well-order of the reals in order type $\omega_1$, then the function $f$ that maps each real to its predecessors violates AS. Observe in this case that the set of pairs $\{(x,y) \mid y\in f(x)\}$ has all vertical sections countable, and all horizonatal sections co-countable, which would violate Fubini's theorem if it were measurable. So it is not measurable.

Conversely, for the direction $\neg AS\to CH$, all violations of AS have essentially this form. To see this, suppose that $f$ is a function without the symmetry property, so that for any two reals $x$ and $y$, either $x\in f(y)$ or $y\in f(x)$. For any real $x$, let $A_x$ be the closure of $x$ under $f$, obtained by iteratively applying $f$ to $x$ and to any real in $f(x)$, and so on to all those reals iteratively. Thus, $A_x$ is a countable set of reals and closed under $f$. Define a relation $y\leq x$ if $y\in A_x$. This is a reflexive transitive relation. The symmetry assumption on $f$ exactly ensures that this relation is a linear relations, so that either $x\leq y$ or $y\leq x$ for any two reals. So it is a linear pre-order. Furthermore, all proper initial segments of the pre-order are countable, since any such initial segment is contained in some $A_y$. In other words, the relation $\leq$ is an $\omega_1$-like linear pre-order of the reals. This implies CH, since the cofinality of this order can be at most $\omega_1$, for otherwise there would be an uncountable initial segment, and so $\mathbb{R}$ would be an $\omega_1$-union of countable sets. That is, the argument shows that every counterexample to AS arises essentially the same way as in your CH argument, but using a pre-order instead of a well-order.

Note that the set $A=\{(x,y)\mid y\in A_x\}$ is non-measurable by the same Fubini argument: all the vertical slices are countable, and all horizontal slices co-countable.

My view is that any philosophical, pre-reflection or intuitive concept of probability will have a very fundamental problem in dealing with subsets of the plane for which all vertical sections are countable and all horizontal sections are co-countable. For such a set, from one direction it looks very big, and from another direction it looks very small, but our intuitive concept is surely that rotating a set shouldn't affect our judgement of its size.

Actually, in regards to your question B), there is a large cardinal axiom that implies $AS$. Your link to the wikipedia article regarding Freiling's axiom of symmetry states the following, in the section "Objections to Freiling's Argument":

"The naive probabilistic intuition used by Freiling tacitly assumes that there is a well-behaved way to associate a probability to any subset of the reals."

This is important, because the truth or falsity of $CH$ is intimately connected with the ability to assign to each subset of the reals, a probability measure.

Consider the following definition, from the Wikipedia article on measurable cardinals:

"A cardinal $\kappa$ is real-valued measurable iff there is a $\kappa$-additive probability measure on the power set of $\kappa$ which vanishes on singletons" (i.e. singletons have probabiity measure zero).

Axiom. Let $\mathfrak c$ be the cardinality of the continuum. $\mathfrak c$ is real-valued measurable.

Consider also the following equivalences from the same wikipedia article:

"A real-valued measurable cardinal ='$\mathfrak c$' exists iff there is a countably additive extension of the Lebesgue measure to all sets of reals iff there is an atomless probability measure on $\mathscr P$($\mathfrak c$).

Note also that the wikipedia article on Freiling's axiom of symmetry linked to your question states that $AS$ is equivalent to $\lnot$$CH$ by a theorem of Sierpinski, and also states that back in 1929, Banach and Kuratowski proved that $CH$ implies that $\mathfrak c$ is not real-valued measurable.

So consider the contrapositive of that statement, that if $\mathfrak c$ is real-valued measurable, then $\lnot$$CH$. Since $AS$ is equivalent to $\lnot$$CH$, then "$\mathfrak c$ is real-valued measurable" immediately implies $AS$. By the definition of real-valued measurable and the aforementioned equivalences found in the wikipedia article on measurable cardinals, Freiling's prereflective probabilistic argument seems to be essentially correct.

This is further confirmed by Noa Goldring in his paper "Measures, Back and Forth Between point Sets and Large Sets", (The Bulletin of Symbolic Logic, Vol. 1, Number 2, June 1995, pp. 182-183 footnotes 17 and 18--also ibid, pp. 171-188.