Set of cardinality continuum contains a perfect set I want to make sure if my claim is true or false :
Every subset of $\mathbb{R}$ of cardinality continuum contains a perfect set
A perfect set is a closed set with no isolated point. It might be there is a counter-example but I did not see it or claim needs to be proved. Any help will appreciated  greatly.
 A: (I assume that "set" here means "set of real numbers.")
Perhaps surprisingly the answer is no, at least assuming the axiom of choice. In fact there are very extreme counterexamples!
A Bernstein set is a set $B$ such that for every perfect set $P$, both $B\cap P$ and $P\setminus B$ are nonempty. Bernstein sets can be constructed via transfinite recursion (this is a good exercise). And we can go further and construct Bernstein sets with various additional properties (see e.g. here). Bernstein sets are one of the classical types of pathological sets of reals under $\mathsf{AC}$; you've probably already seen Vitali sets, and the other two most famous types are Luzin and Sierpinski sets.
(Why must a Bernstein set have size continuum? Well, simply consider any size-continuum family of pairwise-disjoint perfect sets. Any Bernstein set has to intersect each of them, so must have size continuum.)
However, over $\mathsf{ZF}$ alone we cannot prove that Bernstein sets exist, and $\mathsf{ZF+AD}$ proves that every uncountable set of reals does indeed contain a perfect set (and so proves a version of the continuum hypothesis - although it also proves a version of the negation of the continuum hypothesis, namely that there is no injection from $\mathbb{R}$ to $\omega_1$).
