$\newcommand\R{\Bbb R}$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", 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." So, a set $S$ in question, if it exists, cannot be Borel.