Is it still impossible to partition the plane into Jordan curves without choice? 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?
 A: 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 $C_i$. 
A: Isn't the "proper" 3D analogue of filling the plane with closed curves to fill space with surfaces?
I mean, the decomposition of the 3 sphere into two tori (Hopf fibration) is nice and all, but you're using a result involving link theory as the analogue of a result from measure theory...
What I would consider the proper analogue of partitioning the plane with Jordan curves is to partition ${\mathbb R}^3$ with closed, orientable surfaces. And what makes that question superficially a lot more interesting than the ${\mathbb R}^2$ case is that all Jordan curves are homeomorphic to a circle, but closed orientable surfaces can have any genus. However, the proof you give still goes through in that case because a closed orientable surface has nonempty interior...
