converse to the jordan curve theorem Suppose $K\subset \mathbb{R}^2$ is compact and locally connected, and does not contain 0.
Let $A$ be the component of $\mathbb{R}^2-K$ containing 0, and let $B$ be the unbounded component. Assume that $A\neq B$.
If $\partial A = \partial B$, does it follow that $K$ is a Jordan curve?
As a first step, it even true that $\partial A= \partial B=K$?
 A: This is indeed true but nontrivial, it uses some advanced complex analysis and I am unaware of any purely topological proofs. First of all, the assumption $\partial A= \partial B$ and connectivity of $A$ and $B$  imply that $A$ is simply connected. (The same holds for $B'= B\cup \{\infty\}$.) Now, since $K\ne \emptyset$, $A$ is a proper open simply connected subset of the complex plane. Therefore, by the Riemann mapping theorem, there exists a conformal diffeomorphism $f: D\to A$, where $D$ is the open unit disk. 
We will need 
Theorem. (Caratheodory-Torhorst extension theorem) The following are equivalent for a bounded simply connected domain $A\subset {\mathbb C}$:


*

*$\partial A$ is locally connected. 

*${\mathbb C} - A$ is locally connected. 

*The Riemann mapping $f: D\to A$ extends continuously to the boundary circle $S^1$ of the open unit disk $D$. 
You can find a self-contained proof for instance, 
here.  
In our situation, ${\mathbb C} -A$ is the union of $K$ and of 
open subsets of ${\mathbb C}$ (the components of ${\mathbb C} -A$ different from $A$). Therefore, local connectedness of $K$ implies local connectedness of ${\mathbb C} -A$ and, hence, the property that the conformal mapping $f: D\to A$ extends continuously to the unit circle $S^1$. 
I will denote this continuous extension $F: cl(D)\to cl(A)$. I claim that $F$ restricted to $S^1=\partial D$ is 1-1. Suppose not. Let $p, q\in S^1$ be distinct points such that $F(p)=F(q)$. Let $c=pq\subset cl(D)$ denote the chord connecting $p$ and $q$. Its image $C=F(c)$ is clearly homeomorphic to $S^1$ (since $F$ restricted to $c$ is 1-1 except for the endpoints of the chord). Thus, $C$ is a simple topological loop in the complex plane. By the Jordan separation theorem, it separates ${\mathbb C}$ in two components, bounded and unbounded. Since a continuous extension of a conformal mapping cannot be constant on a nondegenerate subarc of the boundary circle $S^1$ (Rado's theorem), there are some points $z$ of $S^1$ such that $w=F(z)$ belongs to the bounded component of ${\mathbb C} - C$. Such points $w$ cannot be reached by an arc $\beta\subset B'=B \cup \{\infty\}$, where the other end of $\beta$ is $\infty$ (since such $\beta$ would have to cross $C$). However, we can also apply Caratheodory's theorem to the simply connected  domain  $B'\subset S^2$. Since $\partial B'= \partial A$ and the latter is locally connected, the same Caratheodory-Torhorst theorem implies that for each boundary point $u\in \partial B'$ there exists a simple arc $\beta\subset B'$, $\beta: [0,1)\to B'$ such that $\lim_{t\to 1} \beta(t)=u$. (Using this theorem is a bit cheating here since existence of such an arc is an application of Caratheodory's theory of prime ends which is used in the proof of the Caratheodory-Torhorst extension theorem.)  Therefore, $F$ is 1-1. But then $F$ restricted to $S^1$ is a homeomorphism to its image (since $S^1$ is compact and the range of the map is Hausdorff). Since $F(S^1)=K$, it follows that $K$ is a Jordan circle. qed
Remark. Note that this proof does not assume that ${\mathbb C} - K$ consists of exactly two components (this is a corollary of the result); it also does not, a priori, assume that $\partial A=K$, this again is a corollary. One can use the three lakes of Wada example to construct a compact connected subset $K\subset {\mathbb C}$ such that two complementary components $A, B$  satisfy $\partial A= \partial B$, but $\partial A\ne K$. Namely, if $A, B, C$ are three lakes of Wada, take $K=cl(C)$. Then $K$ has nonempty interior and, hence, is different from $A$. But in this example $K$ is not locally connected. 
Addendum: Here is how to find points $z\in S^1$ such that $F(z)$ lies in the open disk in ${\mathbb C}$ bounded by the loop $C$. First, the chord $c$ cuts the disk $D$ in two (open) components, $D_1, D_2$; let $\delta_1, \delta_2\subset S^1$ denote the circular arcs equal $cl(D_i)\cap S^1$. The map $F$ sends one of the components, say, $D_1$ outside $C$ and the other one, $D_2$, inside $C$. (This is because if $t$ is a segment in $D$ transversal to $c$, the arc $F(t)$ crosses the circle $C$ in exactly one point and the intersection is transversal.) Now, take any point $u\in D_2$. Since $F(\delta_2)\ne \{q\}$ (as $F$ is not constant on $\delta_2$), there exists a point $z\in \delta_2$ such that $F(z)=w\ne q$. The line segment $uz\subset cl(D)$ is disjoint from $F^{-1}(C)$, hence, $F(uz)$ lies in the same component of ${\mathbb C} - C$ as the point $f(u)$, namely, in the bounded component. Therefore, $w$ lies in the open disk in ${\mathbb C}$ bounded by the Jordan curve $C$.   
