It seems a natural candidate would be to start with CH and perform a countable-support $\omega_2$-iteration of Mathias forcing. Does this achieve your forcing axiom?

Mathias forcing is proper, and it has the pure decision property, meaning that you can decide any statement by shrinking the infinite set only.

Perhaps it helps for others to mention that Mathias forcing has conditions $(s,A)$ where $s$ is a finite stem (a finite sequence of natural numbers) and $A\subseteq\mathbb{N}$ is infinite, with the order placing $(s,A)\leq(t,B)$ when $s$ extends $t$ inside $B$ and $A\subseteq B$.

@AsafKaragila, the infinite sets in conditions can be disjoint, so not compatible even when stem is the same. So one can make an antichain from any almost disjoint family, so this is not ccc.

But even without that, it seems you can force this MA-style with a countable-support iteration and bookkeeping. Is that right?

If CH holds this is equivalent to MA_Cohen, since the poset must be countable. This has no large cardinal strength.

@AlessandroCodenotti Indeed, I asked exactly that question here on MathOverflow 12 years ago: mathoverflow.net/q/93601/1946. Can we prove there is no Borel partition of space into unit circles? And similarly for the other kinds of partitions. It still has no answer.

@RyanBudney Sorry, I don't really know what that means. Could you post an answer explaining the argument?

Please help me retag if appropriate.

Oh, I see, in the weak case the referee is making choices, but there is no assumption of referee choice function, but rather only valid plays, which will be OK by DC. The referee choices only be made during the course of the plays, considered separately.

Can you clarify the role of the referee in the weak quasi-strategy case. I'm unsure what it means to have a play in conformance for the weak case as opposed to the quasi-strategy case. For example, what if there can be no referee making those choices? Then there are no plays in conformance, so the quasi-strategy is vacuously winning?