The Jordan curve theorem can be stated as follows:
Let $p,q\in\mathbb R^2$, $p\ne q$ and $a,b$ be arcs between $p$ and $q$ intersecting only in the endpoints. Then $a\cup b$ divides $\mathbb R^2$ into $2$ connected components, each of which has $a\cup b$ as a boundary.
There's an obvious generalization for three arcs which I think should follow from this, but I'm unable to prove that:
Let $p,q\in\mathbb R^2$, $p\ne q$ and $a,b,c$ be arcs between $p$ and $q$ intersecting only in the endpoints. Then $a\cup b\cup c$ divides $\mathbb R^2$ into $3$ connected components with boundaries $a\cup b$, $b\cup c$, $c\cup b$.
Is it really supposed to be that hard or am I missing something obvious?
What I have so far:
Let $A=\operatorname{int} (b\cup c)$, $B=\operatorname{int} (c\cup a)$, $C=\operatorname{int} (a\cup b)$ be the bounded components,
$X=\operatorname{ext} (b\cup c)$, $Y=\operatorname{ext} (c\cup a)$, $Z=\operatorname{ext} (a\cup b)$ the unbounded components and
$a^\circ=a\setminus\{p,q\}$, $b^\circ=b\setminus\{p,q\}$, $c^\circ=c\setminus\{p,q\}$ the intermediate points of the arcs.
By final components I mean the components of $\mathbb R^2\setminus\{a\cup b\cup c\}$.
Assuming $c^\circ\subseteq C$, we see that $Z$ is the final unbounded component ($Z$ can only get smaller as we add $c^\circ$ to the separating set $a\cup b$, but it won't, because $Z\cap c^\circ\subseteq Z\cap C=0$).
If $a\subseteq A$, then by the same reasoning $X$ is also the final unbounded component, but that's a contradiction, because $\operatorname{bd} Z=a\cup b\ne b\cup c=\operatorname{bd}X$. Therefore $a\subseteq X$ and $A$ is a final component (we add $a$ outside of $A$). Similarly $B$ is final.
So I have $3$ components with the desired properties, but why can't there be a fourth component? (ie. why $A\cup c^\circ\cup B = C$, I can only prove $\subseteq$)
Additionally I assumed $c^\circ\subset C$, I can do that if I use inversion, but I'd be happier I had a proof without that, so I have an additional question:
Is there a simple proof why $a^\circ\subseteq X$, $b^\circ\subseteq Y$, $c^\circ\subseteq Z$ can't happen?
I actually can prove the whole thing using the Jordan–Schönflies theorem (straightening one of the circles and some sphere projections), but there must be a simpler way just from the Jordan's curve theorem (at least I hope).