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?