In his book, Rolfsen partially proves the following equivalent theorem:
Th: If $L$ is a closed subset of $\mathbb R^2$ which is homeomorphic with $\mathbb R^1$, then $\mathbb R^2-L$ has two components, and $L$ is the boundary of each.
He proves that $\mathbb R^2-L$ is disconnected, hence has at least two connected components, and leaves the proof of the fact that it has exactly two components and that $L$ is the common boundary of each as an exercise.
Since a space $X$ is locally connected if and only if the components of each open subset of $X$ are open in $X$, hence each component $C_i$ of $\mathbb R^2-L$ is open in $\mathbb R$. Also, since each $C_i$ is closed in $\mathbb R^2-L$ its limit points not inside itself lie in $L$. Thus $\partial C_i\subset L$. I need hints for the reverse inclusion and to prove that there are exactly two components. Most of the proofs I've seen of these require concepts that were developed specifically for this proof. My question is are there any simple ways of proving these that don't require any constructions that Rolfsen hasn't used before.
Proof is here on page 13.
Let us first prove the following supplement of the JCT.
Lemma. One of the components of $\mathbb R^2 \setminus J$ is bounded and the other component is unbounded. If $C$ denotes the bounded component, then the closure $\overline C = C \cup J$ is compact.
Proof. Let $C_1, C_2$ denote the components. Since $J$ is compact, it is contained in an open disk $D_R(0) = \{x \in \mathbb R^2 \mid \lvert x \rVert < R \}$. The complement $C_R(0) = \{x \in \mathbb R^2 \mid \lvert x \rVert \ge R \}$ is connected, thus it is contained in one of the $C_i$, w.l.o.g. $C_R(0) \subset C_1$. Hence $C_1$ is unbounded. We have $C_2 \subset (\mathbb R^2 \setminus J) \setminus C_1 \subset \mathbb R^2 \setminus C_R(0) = D_R(0)$, hence $C_2$ is bounded. $\overline C_2$ is compact because it is closed and bounded.
Let us next prove that the JCT is equivalent to the "spherical" JCT
SJCT. If $J$ is a simple closed curve in $S^2$, then $S^2 \setminus J$ has two components, and $J$ is the boundary of each.
As a preparation we recall the stereographic projection of $S^2$ from its north pole $N =(0,0,1)$ to the plane.
For any subset $M \subset S^2$ containing $N$ we set $M_* = M \setminus \{N\}$. Stereographic projection from $N$ is the homeomorphism given by $$s : S^2_* \to \mathbb R^2, s(x_1,x_2,x_3) = \frac{1}{1-x_3}(x_1,x_2) .$$
JCT implies SJCT:
Let $J$ be a Jordan curve in $S^2$. W.lo.g. we may assume that $N \notin J$. Otherwise we take a rotation $\rho$ of $S^2$ mapping a point $p \in S^2 \setminus J$ to $N$. The image $\rho(J)$ is again a Jordan curve and if we can prove the SJCT for $\rho(J)$, then SJCT is also true for $J$ since $\rho$ is a homeomorphism.
$s(J)$ is a Jordan curve in $\mathbb R^2$, thus $S^2 \setminus s(J)$ has two components, and $s(J)$ is the boundary of each. Let $C_1$ be the bounded component and $C_2$ the unbounded component. Then $D_i = s^{-1}(C_i)$ are the components of $S^2_* \setminus J$ and $J$ is the boundary of both. Since $\overline C_1$ is compact, also $\overline D_1 = s^{-1}(\overline C_1)$ is compact. Thus $U = S^2 \setminus \overline D_1$ is an open neighborhood of $N$ in $S^2$ and by construction $U_* = D_2$. Since $U$ contains a connected open neighborhood $V$ of $N$ in $S^2$, we see that $D'_2 = D_2 \cup \{N\} = D_2 \cup V$ is connected and $D_1, D'_2$ are the components desired in the SJCT.
SCJT implies JCT:
This is essentially the same argument: If $J$ is a Jordan curve in $\mathbb R^2$, then $K = s^{-1}(J)$ is a Jordan curve in $S^2$. Its complement has two components $D_i$, and $K$ is the boundary of each. Let $N \in D_1$. Then $(D_1)_*$ is still connected and $C_1 = s((D_1)_*)$ and $C_2 = s(D_2)$ are the desired components of $\mathbb R^2 \setminus J$.
Let us finally prove that your Theorem 2 is equivalent to the SJCT.
Theorem 2 implies SJCT:
Let $J$ be a Jordan curve in $S^2$. W.l.o.g. we may assume that $N \in J$. Then $L = s(J_*)$ is a closed subset of $\mathbb R^2$ which is homeomorphic to $\mathbb R$. Theorem 2 shows that $\mathbb R^2 \setminus L$ has two components $C_i$, and $L$ is the boundary of each. The sets $D_i = s^{-1}(D_i)$ are the components of $s^{-1}(\mathbb R^2 \setminus L) = S^2_* \setminus J_* = S^2 \setminus J$. The boundaries $\partial D_i$ contain $J_N$ so that $\partial D_i = J$ since boundaries are closed.
SJCT implies Theorem 2:
Let $L \subset \mathbb R^2$ be closed and homeomorphic to $\mathbb R$. Then $J = s^{-1}(L) \cup \{N\}$ is homeomorphic to $S^1$, thus it is a Jordan curve in $S^2$. Let $D_i$ be the two components of $S^2 \setminus J$. Then the $C_i = s(D_i)$ are the desired components of $\mathbb R^2 \setminus L$.