Here is an answer to your question in another post. Is it possible to partition $\mathbb R^3$ into unit circles? Unfortunately it's not a constructive solution, so we're still missing that.

I do know a nice constructive solution using circles of unbounded radii.

Consider the points on the $x$ axis $(4n + 1, 0, 0)$ for $n \in Z$. Consider the circles of radius $1$ on the $x-y$ plane with center at these points.

Now consider all spheres with radius $r>0$ and center $(0,0,0)$. Each such sphere intersects the set of these circles at exactly two points. Remove these two points from the sphere. What remains of the sphere can certainly be decomposed into disjoint circles. Altogether we have a partition of $R^3$ into a disjoint union of circles.

This solution was pointed out to me many years ago by Yuval Peres.

Here is an answer to your question in another post. Is it possible to partition $\mathbb R^3$ into unit circles? Unfortunately it's not a constructive solution, so we're still missing that.

I do know a nice constructive solution using circles of unbounded radii.

Consider the points on the $x$ axis $(4n + 1, 0, 0)$ for $n \in Z$. Consider the circles of radius $1$ on the $x-y$ plane with center at these points.

Now consider all spheres with radius $r>0$ and center $(0,0,0)$. Each such sphere intersects the set of these circles at exactly two points. Remove these two points from the sphere. What remains of the sphere can certainly be decomposed into disjoint circles. Altogether we have a partition of $R^3$ into a disjoint union of circles.

This solution was pointed out to me many years ago by Yuval Peres.

Here is an answer to your question in another post. Is it possible to partition $\mathbb R^3$ into unit circles? Unfortunately it's not a constructive solution, so we're still missing that.

I do know a nice constructive solution using circles of unbounded radii.

Consider the points on the $x$ axis $(2n + 1, 0, 0)$ for $n \in Z$. Consider the circles of radius $1$ at these points.

Now consider all spheres with radius $r>0$ and center $(0,0,0)$. Each such sphere intersects the set of these circles at exactly two points. Remove these two points from the sphere. What remains can certainly be decomposed into disjoint circles. Altogether we have a partition of $R^3$ into a disjoint union of circles.

This solution was pointed out to me many years ago by Yuval Peres.

Here is an answer to your question in another post. Is it possible to partition $\mathbb R^3$ into unit circles? Unfortunately it's not a constructive solution, so we're still missing that.

I do know a nice constructive solution using circles of unbounded radii.

Consider the points on the $x$ axis $(4n + 1, 0, 0)$ for $n \in Z$. Consider the circles of radius $1$ on the $x-y$ plane with center at these points.

Now consider all spheres with radius $r>0$ and center $(0,0,0)$. Each such sphere intersects the set of these circles at exactly two points. Remove these two points from the sphere. What remains of the sphere can certainly be decomposed into disjoint circles. Altogether we have a partition of $R^3$ into a disjoint union of circles.

This solution was pointed out to me many years ago by Yuval Peres.