The perfect set property says that every uncountable set of reals contains a perfect subset.
Now consider the following statement:
P: For every $X\subset\mathbb{R}$, either $X$ or $\mathbb{R}\setminus X$ contains a (non-empty) perfect set.
Clearly, the perfect set property implies the second one. I've heard that the converse is false, but I couldn't find a model of ZF+P (this would obviously entail a failure of choice). Moreover, is it possible to get a model where P holds without assuming the existence of an inaccessible?
Any references for this? I've looked at Jech, Rubin, this paper , and Googled a bunch, but found nothing.
Yes. This was studied by John Truss in his paper that you mention,
Specker's theorem tells us that the perfect set property implies $\omega_1$ is a limit cardinal in $L$, but without countable choice, $\omega_1$ can be singular. And Truss shows that this is consistent with the perfect set property.
Moreover, he shows, in a fairly simple family of models every set of reals can either be well-ordered or it contains a perfect set (and the reals cannot be well-ordered), which is an intermediate statement between your $\rm P$ and the perfect set property. This is even consistent with Dependent Choice to some fixed level.
So now take a set of reals, if it can be well-ordered, then its complement cannot be, and thus contains a perfect set. If your set cannot be well-ordered, then it contains a perfect set.