On p441 of Real Mathematical Analysis by Pugh, but also in this question and this blog post, the axiom of choice is applied to a set of irrational orbits $O(x)=\{R^k(x):k\in \mathbb{Z}\}$ to create a non-measurable set. The rotations are applied to the unit circle represented as $[0,1)$.
From what I understand, this means selecting a value $k$ for each point $x$ and applying the irrational rotation $R$ $k$ times to get a point of the orbit. But if I select $k=1$ for every $x$, I just get one rotation applied to the unit circle and the unit circle itself as the result, which is obviously measurable. Yet, this is a set that contains exactly one point from every orbit as required. I'm probably missing something very simple here so I'd love to find out what that is. Thanks!
The point you're missing is that even for $k=1$, you're not getting the circle back. For example, $x$ and $R(x)$ both lie in the same orbit. So you can only choose one of them if you're choosing a single point from each orbit.
Another way to look at this is to simply consider the partition into the orbits. Once you pick a point in the orbit, you can call that $x$, and then every other point is $R^k(x)$ for some $k\in\Bbb Z$. But before we decided which $x$, there's no reason to prefer one over the other.