Can Two Compact, Non-Separating Subsets of the Plane be Separated by Jordan Curves? Suppose $X, Y \subset \mathbb{R}^2$ are compact, mutually disjoint, and neither separates the plane.
How does one prove that there are disjoint Jordan curves $J_1, J_2$ such that $X$ is in the interior region of $J_1$, $Y$ is in the interior region of $J_2$, and neither of $J_1, J_2$ is contained in the interior region of the other?
The requirements on $X$ and $Y$ are necessary; for example, take two disjoint circles, one inside the other.  Or consider the graph of $|\frac{1}{x}|$.  The above theorem can be proven using the classification of non-compact surfaces, but that's too big a cannon for my purposes; I want something elementary.  Schoenflies, JCT, Zoretti etc. is fine, but nothing too beyond that!
If $X$ or $Y$ is connected then it's known due to either some algebraic topology results or Zoretti's Theorem, but in my case I'll be dealing with Cantor sets; I am concocting a proof of the Denjoy-Riesz Theorem that's actually readable and have put one together, but there's a somewhat tedious step that I'd like to simplify before I post it, if possible.
Thanks a lot!  The specific case that if $X, Y$ are Cantor Sets is also of interest; that's the tedious step in my Denjoy-Riesz proof.  I proved it using finite covers of open balls, some geometric arguments about these piecewise-nice neighborhoods, then further subdividing and filling in to avoid one neighborhood 'wrapping around' the neighborhood of the other Cantor set.  It's not an aesthetic proof.
 A: Here is a sketch how can do it. Drawing a picture will be helpful.
$X,Y$ have distance $d > 0$. Therefore we can cover $X,Y$ by disjoint compact subpolyhedra $P, Q$ of $\mathbb R^2$ (e.g. unions of sufficiently small squares). Next we take disjoint regular neigborhoods $M, N$ of $P, Q$. These are compact PL manifolds with boundary. Note that the boundaries of $M, N$ are disjoint from $X, Y$ since $X,Y$ are contained in the manifold interiors $\mathring{M}, \mathring{N}$.
Let us first consider $M$. It has only finitely many components; their interiors cover $X$. Let $C$ be one them; it is a compact PL manifold with boundary. Then $X_C = X \cap C$ is compact and does not intersect $\partial C \subset \partial M$.
We shall cut off pieces of $C$ to obtain a PL copy $D_C$ of the disk $D^2$ such that $X_C \subset D_C \subset C$ and $X \cap \partial D_C = \emptyset$.
It is known that $C$ has the form $C = D^* \setminus \bigcup_{i=1}^n \mathring D_i$ where
$D^*, D_1, \ldots, D_n$ are PL disks such that $D_i \subset \mathring D^*$ for $i=1,\ldots, n$ and $D_i \cap D_j = \emptyset$ for $i \ne j$. It is of course possible that $n = 0$, i.e. $C = D^*$, but this is the trivial case. So let $n > 0$. Pick $x^* \in \partial D^*$ and $x_1 \in \partial D_1$. Since $X$ does not disconnect $\mathbb R^2$, there is a path $u$ from $x^*$ to $x_1$ which avoids $X$. Thus the distance between $u(I)$ and $X$ is positive. There is a PL path $v$ from $x^*$ to $x_1$ which is so close to $u$ that it also avoids $X$. We can modify $v$ as follows: If $v$ enters some $D_j$ at time $t$ and leaves it at time $s > t$, we can replace $v$ on $[t,s]$ by a PL path in $\partial D_j$. Similarly, if $v$ enters $\mathbb R^2 \setminus \mathring D^*$ and leaves it at time $s > t$, we can replace $v$ on $[t,s]$ by a PL path in $\partial D^*$. Thus we may assume w.l.o.g. that $v$ is a path $C$ which avoids $X$. As the next step we choose a PL path $w$ in $C$ from $x^*$ to $X_1$ which so close to $v$ that it avoids $X$ and has the property $w((0,1)) \subset \mathring C$ ("pushing $v$ away from the boundary''). W.l.o.g. we may assume thet $w$ is a PL embedding (if it is not, we remove possible loops between points of self-intersection). Now choose a regular neigborhood $W$ ot $w(I)$ in $C$ which is so small that it avoids $X$ ("thickening of $w$ to a strip connecting $\partial D^*$ and $\partial D_1$''). There is a PL homeomorphism $h : I \times I \to W$ such that $h(\{0\} \times I) \subset \partial D^*$, $h(\{1\} \times I) \subset \partial D_1$ and $h((0,1) \times I) \subset \mathring C$. Then $C' =  C \setminus h(I \times (0,1))$ is again a PL manifold neigborhood of $X$ having the form $C' = D_1^* \setminus \bigcup_{i=2}^n \mathring D_i$ with a PL disk $D^*_1$. Proceeding inductively we end with a PL disk $D$  such that $X_C \subset D_C \subset C$ and $X \cap \partial D_C = \emptyset$.
Now let $C_1,\ldots, C_m$ be the components of $M$ and assume that they are reduced to disks as above. They cover $X$ and their boundaries do not intersect $X$. Similarly let $C'_1,\ldots, C'_r$ be the components of $N$ and assume that they are reduced to disks. By a similar approach as above we can find a a strip $W_1$ connecting $\partial C_1$ and $\partial C_2$ and avoiding all other $C_i$ and $C'_j$. Then $\bar C_2 = C_1 \cup W_1 \cup C_2$ is a PL disk whose boundary does not intersect $X$.  Proceeding inductively we end with a PL disk $\bar C_m$ containing $X$ in its interior and being disjoint from $N$. Doing the same with the $C'_j$ results in a PL disk $\bar C'_r$ containing $Y$ in its interior and being disjoint from $\bar C_m$. The boundaries of $\bar C_m$ and $\bar C'_r$ are the desired Jordan curves.
