Does every uncountable Borel subset of $\mathbb R$ contains a perfect subset？ This question came from (London Mathematical Society Student Texts) Krzysztof Ciesielski-Set Theory for the Working Mathematician-Cambridge University Press. Chapter 6.2 Exercise 5. 
I have thought about it for a few weeks and asked some friends. 
The extension question is whether every Borel subset of $\mathbb R$ is either countable, or contains a perfect subset？
 A: Yes - this is called the perfect set property. The analytic (= continuous image of Borel) sets also have this property; for more complicated sets (e.g. complements of analytic sets), the perfect set question is undecidable from the usual axioms of set theory.
While it's trivially true for open sets, already for closed sets it takes some work (the easiest approach, given a closed set $C$, is to consider the set of elements of $C$ around which $C$ is "locally uncountable" - that is, those $c\in C$ such that every open $U$ containing $c$ has uncountable intersection with $C$).
The full result follows from Borel determinacy, which is a very hard theorem; off the top of my head I don't know a proof that doesn't use this. Kechris' book Descriptive set theory is a very good source (as is Moschovakis' book).
A: I have an idea mainly from Noah Schweber's advice. I don't know is it exactly right or not?
The simple idea is a subset of a Polish space (like the real numbers) has the perfect set property if it is either countable or has a nonempty perfect subset (Kechris 1995, p. 150).
A proof from Noah Schweber' idea is as follow:
Firstly, it is trivially true for open sets, (since the definition of borel set is from the open set). If a borel subset of R is/or contains an open set, we may let it be $B(p, \epsilon)$，then in $R$ there is a $\delta < \epsilon$, such that $[p-\delta, p+\delta]$ is a close set ，and this closed set has no isolated point，and this set is perfect set. From the definition of perfect set, that is,  in the field of topology, a subset of a topological space is perfect if it is closed and has no isolated points. 
But for closed sets it takes some difficult work. The easiest approach, given a closed set $C$, if it doesn't contain a perfect set ,that is every point/element of $C$,let it be  $c \in C$ is isolated points, but this is impossible. Because every family U of pairwise-disjoint open subsets of R is at most countable. For every isolated point of $C$, we could find an enough "small" open set $B(c, \epsilon)$, to cover the isolated point, and we could ensure all the open sets are pairwise-disjoint. So we have only at most countable isolated points. But this closed set of Borel subset is uncountable. So this closed set contains a part of uncountable closed set.
Then from (Cantor–Bendixson theorem) ,every uncountable, closed subset F of R can be represented as a disjoint union of a perfect set P and an at most countable set C.So we have a perfect set.
