Must the complement of the countable union of disjoint arcs (in the plane) be connected?

A *(Jordan) arc* is the injective image of a continuous function from $[0,1]$ to $\Bbb R^2$. It is known that, if I have *finitely* many pairwise disjoint arcs, the complement of their union must be connected, and even path-connected. However, this no longer is true when we allow ourselves infinitely many arcs (take the union of line segments of the form $\{x\}\times[0,1]$ for $x\in\Bbb R$).

In fact, the "path-connected" part is not even true for countably many arcs!

I know three counterexamples (special thanks to Balarka and Alessandro):

- Let $U_r$ be a closed circular arc of $270$ degrees, with the $90^\circ$ opening facing up, of radius $r$. Let $D_r$ be the same but with the opening facing down. Then the complement of: $$U_{1/2}\cup D_{2/3}\cup U_{3/4}\cup D_{4/5}\cup\dotsb$$ is not path-connected.
- The complement of: $$\bigcup_{n=1}^\infty(\{1/n\}\times[-n,n])$$ is not path-connected. This gives an example made of only straight lines, though it's unbounded.
- Let $S_1=[-1,1]\times\{1\}$ and $S_{-1}=[-1,1]\times\{-1\}$ be the top and bottom sides of a square. Let $T_{-1,n}=\{-1-\frac1n\}\times[-1,1]$ and $T_{1,n}=\{1+\frac1n\}\times[-1,1]$ be sequences of segments approaching the left and right sides of the same square. Then the complement of: $$S_{-1}\cup S_1\cup\bigcup_{n=1}^\infty T_{-1,n}\cup\bigcup_{n=1}^\infty T_{1,n}$$ is not path-connected. This gives an example that is bounded, is made of straight lines, and has all of its lines the same length.

However, while these complements are all not path-connected, they all are *connected*. Hence the question: Must the complement of countably many disjoint arcs be connected?

I've thought about this all day and I haven't been able to solve it. Homology seems useless, since it can only measure path components, not connected components. On the other hand, if we replace $\Bbb R^2$ with other spaces, it can become false (take for example the plane minus open disks of radius $\frac13$ centered at the integers on the $x$-axis), so the proof must use some special property of $\Bbb R^2$.