>$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set? 

Of course, such a set $S$, if it exists, must be of measure $0$.

["The axiom of choice implies the existence of sets of reals that do not have the perfect set property"][1], so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

On the other hand, it ["is well-known that assuming the countable axiom of choice that every
uncountable Borel set contains a [nonempty] perfect set."][2] So, a set $S$ in question, if it exists, cannot be Borel. 


  [1]: https://en.wikipedia.org/wiki/Perfect_set_property
  [2]: https://arxiv.org/abs/0806.1957v1