Recently I came across a post over at MO (https://mathoverflow.net/questions/162324/covering-the-space-by-disjoint-unit-circles) that claimed Sierpinski proved that $\mathbb{R}^2$ cannot be expressed as the disjoint union of circles. I have been unable to find a reference, does anyone have an idea where one could find the details to this result?
EDIT: A general argument is fine, too, but I'm also interested in Sierpinski's original proof.
I think it should go like this. Suppose by contradiction that we can write $\mathbb{R}^2$ as a disjoint union of (nondegenerate) circles. Pick one such circle. It bounds a disk in $\mathbb{R}^2$, so call the closure of that disk $K_1$. Any point in the (nonempty) interior of the disk is on some other circle. By disjointness, that circle is entirely in the interior of $K_1$, so we get another compact set $K_2 \subsetneq K_1$. Actually we can construct an infinite chain of such sets, $K_1 \supsetneq K_2 \supsetneq \dotsb$. Notice that $$\operatorname{diam}(K_n) := \sup\{d(x,y) \mid x, y \in K_n\}$$ is positive, finite, and for $n \ge 2$ we can choose it to be strictly less than $\frac12\operatorname{diam}(K_{n-1})$. If all circles in the interior of $K_{n-1}$ are concentric this is obvious, and if not, by disjointness the sum of the diameters of two nonconcentric circles has to be smaller than $\operatorname{diam}(K_{n-1})$, so one of them has diameter smaller than $\frac 12\operatorname{diam}(K_{n-1})$.
By Cantor's intersection theorem, the intersection $\bigcap_{n=1}^\infty K_n$ is nonempty. I claim it has exactly one point $p$, since $\lim_{n\to\infty}\operatorname{diam}(K_n) = 0$. By the above argument, the circle containing $p$ must be contained in each $K_n$, so it would have to be contained in the intersection. This is a contradiction.