Is it possible to partition $\mathbb R^3$ into unit circles?
The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3=\{p_\alpha:\alpha<\phi\}$ be an enumeration of the points of space. We define a unit circle $C_\alpha$ containing $p_\alpha$ by transfinite recursion on $\alpha$, for some $\alpha$ we do nothing. Here is the recursion step. Assume we have reached step $\alpha$ and some circles $\{C_\beta:\beta<\alpha\}$ have been determined. If some of them contains (=covers) $p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p_\alpha$ that misses all the earlier circles. For that, we first choose a plane through $p_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes through $p_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let $K$ be the plane chosen. The earlier circles intersect $K$ in less than continuum many points, so it suffices to find, in $K$, a unit circle going through $p_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ that pass through $p_\alpha$ and each of the bad points disqualifies only 2 of them.