Unmeasurable dense subset of circle

Question

I need to find unmeasurable dense subset of circle. I think, that I found the unmeasurable set, but I can't show that it is dense. Here is my construction. Let's take $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, consider the irrational rotation of the circle on angle $\alpha$. Let's take orbits of all points on the circle under the rotation, then choose the only one point from every orbit. This is set $X_0$. Then $X_j=X_0+\alpha*j$. So $\forall j$ the set $X_j$ is unmeasurable. I need to find the point of $X_j$ in any neighbourhood of any point on the circle to show that $X_j$ is dense. But I don't know how to do it.

Answer 1

Take any unmeasurable subset $X$ and add a countable dense subset $Y$, that is take $X \cup Y$.

Answer 2

Let $V$ be a Vitali set in $[0,1)$ ( by AC choose a representative for each class of the relation $x\sim y$ iff $x-y\in\mathbb{Q}$) and take its image under the continuous bijection $t \to (\cos(2\pi t, \sin 2\pi t)$. This is as required.