Consider four disjoint points in the plane, $v_{1}$, $v_{2}$, $v_{3}$ and $v_{4}$.
The cycle, $C:=v_1v_2v_{3}v_{4}v_{1}$, is the union of the (Jordan)
arcs, $A_{12}$, $A_{23}$, $A_{34}$, and $A_{41}$, in which every
$A_{ij}$ is an arc between the points $v_{i}$ and $v_{j}$. 
By the Jordan Curve Theorem (JCT), the region enclosed by $C$, has an
interior and an exterior, respectively denoted by $\text{Int}(C)$ and
$\text{Ext}(C)$. 

In addition, we assume that there exists an arc, $A_{13}$, entirely
contained within $\text{Int}(C)$, apart from its endpoints. Clearly,
$A_{13}$ appears to partition $\text{Int}(C)$ into two disjoint
regions. Such a statement may be written as follows, 
\begin{equation}\notag
    \text{Int}(C) = \text{Int}(C_{1231}) \cup \text{Int}(A_{13}) \cup \text{Int}(C_{1341}),
\end{equation}
where the cycles $C_{1231}$ and $C_{1341}$ are defined as
$C_{1jk1}:=v_{1}v_{j}v_{k}v_{1}$, and $\text{Int}(A_{13})$ denotes the
arc $A_{13}$, without its endpoints. 

When attempting to prove such a statement, however, one keeps
switching from the interiors to the exteriors of each sub-region,
without much success. Indeed, the sole use of the JCT
does not appear to be sufficient in this context, since it solely
defines the interior of a simple closed curve, relative to its exterior
and vice-versa, thereby making it seemingly impossible to combine the interiors 
$\text{Int}(C_{1231})$ and $\text{Int}(C_{1341})$, in order to produce
$\text{Int}(C)$. 

Am I missing something obvious? Or is such a fact too elementary to be
proved? Could this be taken to be part of the definition of the faces
of a plane graph, for instance? But if one can prove the JCT, why
wouldn't it be possible to prove such a similar claim?