This results from the following fact: there exists a negligible uncountable Borel subset of . More precisely, its cardinality is that of .
Supposing this is true, then every subset of is a Lebesgue subset (because they're included in a negligible Borel subset). However, because , the set of the subsets (i.e. the power set) of is in bijection with the power set of . This proves what we want.
Coming back to the announced fact, the usual example is a Cantor subset, and I'd bet you've already seen this. For instance, it can be the set of the numbers in such that all of their decimals (in base 10) are even. Or (the usual one) the set of the numbers in such that the only numbers appearing in their representation in base 3 are 0 and 2.