It is an easy exercise to show that the Euclidean plane cannot be partitioned into round circles (note however that it is possible to do so for $\mathbb{R}^3$). It seems almost obvious that it is not possible to partition the plane into Jordan curves either.


However, I am not able to design a proof that does not use the choice axiom. With choice, assume you have such a partition $J$ and define on $J$ a partial ordering: $j<k$ if the curve $j$ is contained in the interior of $k$. Any decreasing chain $j\_n$ has a lower bound: if $K\_n$ is the closure of the interior of $j\_n$, then $K\_n$ is a decreasing sequence of compact sets, thus there is some point $x$ in the intersection. Then the curve of $J$ that contains $x$ is a lower bound of $(j\_n)$. Now Zorn Lemma ensures that there is a minimal element $j$ in $J$. But this is obviously impossible since $j$ would have a non-empty interior, therefore containing another curve of $J$.


The question is therefore the following: can we prove that there exist no partition of the plane into Jordan curves without assuming the Choice axiom?



Is it still impossible to partition the plane into Jordan curves without choice?
It is possible to change your argument so that the choice is over countable set; hope this is good enough. Namely, topology on the plane has countable base (say, circles at rational points with rational radii); let's index this base as $U\_1,\dots,U\_n,\dots$; your argument can be used to construct a sequence of Jordan curves $C\_1,\dots,C\_n,\dots$ such that $C\_{i+1}$ is contained in the interior of $C\_i$, and $U\_i$ is not contained in the
interior of $C\_i$. 


