A major argument against Freiling's Axiom of Symmetry is the following (this from the wikipedia article of the same name):

"The naive probabalistic notion used by Freiling tacitly assumes that there is a well-behaved way to associate a probability to any subset of the reals. But the mathematical formalization of the notion of 'probability' uses the notion of measure, yet the axiom of choice implies the existence of nomeasurable subsets, even of the unit interval..."

However, there seems a way to so formalize Freiling's naive, pre-reflective arguments for associating a probability to any subset of the reals. I propose the following:

Definition. 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 probability measure zero)." (This from the wikipedia article on measurable cardinals.)

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

It is interesting to note that the wikipedia article on measurable cardinals states that back in 1929, Banach and Kuratowski proved that $CH$ implies that $\mathfrak c$ is not real-valued measurable.

Since the Axiom of Symmetry is equivalent to the negation of $CH$, assuming that $\mathfrak c$ is real-valued measurable as an axiom immediately implies that the axiom of symmetry holds.

Also one has from the wikipedia article on measurable cardinals, the following (at least according to my understanding at this moment):

"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 $\mathscr P$($\mathfrak c$).

Question. In what way(s) does "$\mathfrak c$ is real-valued measurable' not (if any) capture Freiling's naive, pre-reflective arguments?

Question. Which large cardinal axioms imply that $\mathfrak c$ is real-valued measurable?

Question. What happens to the axiom of choice in the presence of my proposed axiom?

Question. Why is my proposed axiom yet another example of an axiom that "could not be"?

[Addendum: Two questions in response to Monroe's answer:

Let $S$ be "There is a subset of $\mathscr R^2$ which is countable on every horizontal line and co-countable on every vertical line." It is known that $CH$ implies $S$. One can also denote $S$ by "There exists a Sierpinski example", so one should be able to express the contrapositive as follows:

" 'There is no Sierpinski example' implies $\lnot$$CH$."

Question: Does $ZFC+$$($$\bigstar$$)$$\vdash$$\lnot$$CH$?

Question: Does $ZFC+$$($$\bigstar$$)$$\vdash$"There is, on $\mathscr R^2$, a product measure that is a total measure?"]