Is there a nonmeasurable set in R in which all the measurable subsets are countable? I think because there exist so many uncountable zero measure sets, it ought to be very easy to create a counterexample. However I failed to get an idea about this. 
 A: That is the definition of a Sierpiński set: an uncountable set (or equivalently, a non-measurable set) whose every measurable subset is countable. The existence of a Sierpiński set of real numbers is independent of ZFC (standard set theory including the axiom of choice), but follows from the continuum hypothesis.
To construct a Sierpiński set using the continuum hypothesis, we use the fact that every set of Lebesgue measure zero is contained in a $G_\delta$ set of measure zero. Enumerate all the $G_\delta$ sets (or all the Borel sets) of measure zero in a transfinite sequence $A_\alpha,\ \alpha\lt\omega_1.$ Choose recursively points
$$x_\alpha\in\mathbb R\setminus\left(\{x_\beta:\beta\lt\alpha\}\cup\bigcup_{\beta\lt\alpha}A_\alpha\right);$$
then $X=\{x_\alpha:\alpha\lt\omega_1\}$ is a Sierpiński set.
On the other hand, it is consistent with ZFC (see Martin's axiom) that every set of real numbers of cardinality $\aleph_1$ has Lebesgue measure zero. In that case, there are certainly no Sierpiński sets, seeing as every uncountable set has a subset of cardinality $\aleph_1.$
