Axiom of choice and a set in the plane that intersects every line in two points

In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects each line exactly twice'' (called $2$-point sets [I think]).

I was talking with an students and we were wondering if, in the absence of Choice, this result is still true.

To make my question more precise, I know that we only need Choice to well order the reals and, I believe, you can make the set to be non-measureble but I'm not sure if it is non-measurable itself (I believe that using CH you can make a $2$-point set of measure zero [but, to be honest, I haven't work out the details]).

So, being more concrete, how much choice is need to create such a set? Are there models of ZF without them?

• It seems that this is open whether or not such set can even be Borel, let alone $G_\delta$. In that case, I think you can probably construct one without appealing to the axiom of choice. – Asaf Karagila Dec 3 '15 at 23:14
• Arnie Miller showed that you don't need a well ordering of reals in the sense that there are ZF models with 2-point sets where the set of reals cannot be well ordered: math.wisc.edu/~miller/res/two-pt.pdf He also showed that in $L$, there are conanalytic 2-point sets which is the best known upper bound - An analytic 2-point set is necessarily Borel. Mauldin has results connecting this to geometric measure theory: math.unt.edu/~mauldin/papers/no100.pdf – Ashutosh Dec 4 '15 at 0:13
• On measure and category: There is a 2-point set within $\{z \in \mathbb{R}^2 : |z| \in C\})$ where $C$ is any set of reals which, say, meets every interval on a perfect set . So there is always a meager null 2-point set. – Ashutosh Dec 4 '15 at 0:56
• Great question! This one is closely related: mathoverflow.net/questions/272527/…. – Will Brian Feb 21 at 13:45