Accessible Proof of Denjoy-Riesz Theorem?

Ok, I think the reason that this oft-cited result isn't proven in many books is because a stronger theorem was later proven using means that weren't really any more sophisticated, but the proof is still pretty long. The only place besides Kuratowski that I found this is Moise's "Geometric Topology in Dimensions 2 and 3" where he proves the following, stronger theorem instead:

Let $C_1$ and $C_2$ be Cantor sets in the plane, and let $f: C_1 \rightarrow C_2$ be a homeomorphism. Then there is a homeomorphism $\tilde{f}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ extending $f$.

The proof of this comprises chapters 12 and 13 of his book. Let $I = [0,1]$. I'm lazy, but sometime in the next few days I'll post my simpler proof of the following, ostensibly stronger theorem as another answer:

Let $C_1$ and $C_2$ be Cantor sets in the plane, and let $f: C_1 \rightarrow C_2$ be a homeomorphism. Then there is an ambient isotopy $F: \mathbb{R}^2 \times I \rightarrow \mathbb{R}^2$ sending $C_1$ to $C_2$ which equals $f$ on $C_1$ at time $t=1$.

To be fair to Moise this is implied by other theorems he proves elsewhere in the book, namely that any stable homeomorphism of the plane (i.e. one that leaves some open set in $\mathbb{R}^2$ fixed) is isotopic to the identity. The construction he uses produces a stable homeomorphism - but then you also need the theorem on stable homeomorphisms.

1. Either of these theorems immediately implies the Denjoy-Riesz Theorem, since we can cover the usual middle-thirds Cantor set with an arc and then isotope the arc onto any Cantor set.

2. Either theorem can be restated for compact, totally disconnected sets (equiv. zero-dimensional or totally separated) since each resides in a Cantor set.

3. No such set separates the plane, since no subset of an arc does.

For those who were curious, a different generalization by Moore and Kline goes as follows:

A compact set $X$ in $\mathbb{R}^2$ can be covered by an arc iff each component of $X$ is a point or a closed arc $A$ such that if $a \in A$ is a limit point of $X \setminus A$ then $a$ is an endpoint of $A$.

The proof is nigh-unreadable and makes references to many technical results in other papers. I will try to simplify this one as well, but my technique seems ill-suited to it without something smart. Maybe you can help once you see my method for the isotopy extension version.

I'll use the Schoenflies Theorem: If $J_1$ and $J_2$ are Jordan curves in the plane (or sphere) there is an ambient plane (resp. sphere) isotopy sending $J_1$ to $J_2$. This is sometimes called the "strong" Schoenflies Theorem (it's a consequence of the Annulus Theorem). Let $\text{Int}(J)$ and $\text{Ext}(J)$ denote the bounded and unbounded complementary regions of $J$, respectively.

If $J_1, \dots, J_n$ are a collection of mutually disjoint Jordan curves, then call the collection simple if no $J_i$ is contained in any $\text{Int}(J_k)$. By induction and repeated use of the Schoenflies Theorem:

Observation 1: Any two simple collections of Jordan curves are ambient isotopic if they have the same number of elements.

A continuum (plural continua) is a compact, connected metric space. We need the following theorem: A nested intersection of continua is a continuum, as proved in the body of the question here. Relevant is that since the Cantor Set is totally disconnected, it doesn't contain a non-degenerate continuum.

Let $B_\epsilon(x) = \lbrace y: |x-y| \leq \epsilon \rbrace$ and let $B_\epsilon(X) = \cup_{x \in X} B_\epsilon(x)$. If $X \subset \mathbb{R}^2$ is a continuum then let $X_\circ$ denote the filled in continuum obtained by adding each bounded component of its complement (it's a continuum since its complement is open and a sufficiently large disc around $X$ has the boundary bumping property).

Lemma 1: Let $C_1$ and $C_2$ be disjoint Cantor sets in the plane (or sphere). Then there is a simple pair of disjoint Jordan curves $J_1$ and $J_2$ with $C_i \subset \text{Int}(J_i)$.

Proof: By Observation 1, it's sufficient to show that there are simple collections $\mathcal{A}_1$ and $\mathcal{A}_2$ containing $C_1$ and $C_2$ respectively such that $\mathcal{A} = \mathcal{A}_1 \cup \mathcal{A}_2$ is also simple, since in this case it is explicitly constructible after suitable isotopy. Since each $C_i$ is compact each $B_\epsilon(C_i)$ is a finite cover by Euclidean discs and thus each complementary component is a Jordan domain (apart from the unbounded one).

Letting $\epsilon$ be small enough that $B_\epsilon(C_1) \cap B_\epsilon(C_2) = \varnothing$, we get disjoint covers of each such that filling in their components produces covers by Jordan domains. However, it's possible that for some components $D$ of $B_\epsilon(C_1)$ and $E$ of $B_\epsilon(C_2)$, we have $D_\circ \cap E \neq \varnothing$, without loss of generality. In which case $E$ is contained in a complementary region of $D$ since it's disjoint from it.

In $D$, find a $\delta$ sufficiently small so that $E$ is no longer contained in any complementary component of $B_\delta(D \cap C_1)$. If none such existed, then for $\delta_n = \frac{1}{n}$ we have that $E$ is contained in a bounded complementary component of each $B_{\delta_n}(D \cap C_1)$, say of component $D_n$ respectively which are nested by construction. But this is impossible since it would imply the existence of a non-degenerate continuum in $C_1$, as $D_n$ cannot converge to a point while all separate $E$ from $\infty$ unless $D_n \rightarrow E$, but $D_n \subset D$ which is disjoint from $E$ and closed.

After doing this in each component of $B_\epsilon(C_1)$, by finite descent we have a simple collection $\mathcal{A}_1$ of Jordan curves with interior domains covering $C_1$ such that none intersect $C_2$, nor do their interiors. Doing the same for $B_\epsilon(C_2)$ produces the desired family $\mathcal{A}$. QED

Let $C$ be the standard Cantor ternary set, viewed as a subset of the real line $\mathbb{R} \times \lbrace 0 \rbrace \subset \mathbb{R}^2$. Let $E$ denote the set of endpoints of the (bounded) deleted intervals in the real line, and let $E_n$ denote the endpoints of the intervals with length $\frac{1}{3^n}$, called the $n$th level of $E$. Note that $E$ is dense in $C$. Call the removed intervals $U_j^n$ the $n$th level of removed intervals. Call the $2^n$ neighborhoods $D_j^n$ complementary to the $n$th level of removed intervals the $n$th levels of $C$. Note that the collection $\mathcal{D}_n = \lbrace D_j^n \rbrace$ is a covering of $C$ for every $n$, and each $D_j^n$ is a Cantor set that's clopen in $C$.

Let $I = [0,1]$ be the unit interval. If $X, Y \subset \mathbb{R}^2$ and $f: X \rightarrow Y$ is a homeomorphism, we say that an ambient isotopy $F: \mathbb{R}^2 \times I \rightarrow \mathbb{R}^2$ extends $f$ if we have that $f(X) \equiv F(X \times \lbrace 1 \rbrace)$.

Theorem: If $C_1, C_2$ are Cantor sets in the plane and $f: C_1 \rightarrow C_2$ is a homeomorphism, there exists an ambient plane isotopy extending $f$.

Note that the proof below also works for the sphere.

Proof: Let $g$ be an arbitrary homeomorphism from $C_1$ onto $C$, the standard ternary Cantor set. We produce an ambient isotopy from $C_1$ to $C$ extending $g$. Then there also exists isotopies extending arbitrary homeomorphisms $C_2 \rightarrow C$, and homeomorphisms between compact metric spaces are quotient maps. Thus we may factor $f$ through $g$ to obtain an $h^{-1}$ such that $f = h^{-1} \circ g: C_1 \rightarrow C_2$, a concatenation of ambient isotopies and thus an ambient isotopy itself.

By Lemma 1 and Observation 1 there is a simple pair of Jordan curves $J_1, J_2$ around $g^{-1}(D_1^1)$ and $g^{-1}(D_2^1)$, and these may be isotoped to narrow discs sitting above $D_1, D_2$ as shown below such that the preimages of the first level of $E$ share $x-$coordinates with their images. The interior of each is homeomorphic to the plane, so by induction we may apply Lemma 1 and Observation 1 repeatedly to isotope inverses as necessary to approximate $g$ on $g^{-1}(E)$. By concatenating isotopies which differ on uniformly smaller sets, we obtain a homotopy on the plane sending $C_1$ to $I$ and which is a homeomorphism for all $t < 1$. We need to show it's a homeomorphism at $t = 1$ on $C_1$, as it is clearly bijective off of $C_1$.

Since $g$ is a homeomorphism, $g^{-1}(E)$ is dense in $C_1$ so it uniquely determines the function $F$ restricted at time $t = 1$. Since both are compact metric spaces it's sufficient to show that $F$ is bijective on $C_1$ at $t = 1$. But this is immediate by construction, since any two points in $C_1$ are contained in disjoint elements of some $n$th level of $C$ and these have images with distance bounded below by $\frac{1}{3^n}$. QED

In higher dimensions the proof breaks down at the observation.

Anyone see how to continue to grab the Moore-Kline result?