Classify, except isomorphism, all the covering spaces on $\mathbb{S}^1$ with three branches such that the total space is not necessarily arc-connected.
I know that $\mathbb{Z}$ acts continuously and properly discontinuous in $\mathbb{R}$, so that $p:\mathbb{R}\to \mathbb{R}/\mathbb{Z}\cong\mathbb{S}^1$ is the universal covering space, then the subgroups of index three of $\mathbb{Z}$ correspond to three-sheet covering spaces on $\mathbb{S}^1$, I affirm that $3\mathbb{Z}$ has index three in $\mathbb{Z}$ and thus $r: \mathbb{R}/3\mathbb{Z}\to \mathbb{R}/\mathbb{Z}\cong\mathbb{S}^1$ is a three-sheet covering space. There is more? I think that the disjoint union of any three covering spaces is also a three-sheet covering space, for example $\mathbb{R}\sqcup \mathbb{S}^1\sqcup\mathbb{R}$, $\mathbb{S}^1\sqcup\mathbb{S}^1\sqcup\mathbb{S}^1$, am I thinking correctly? Thank you very much.
If you rule out "$\mathbb{R} \sqcup \mathbb{S}^1 \sqcup \mathbb{R}$" as @tomasz points out you should do, then there's still one missing. Namely, $\mathbb{R}/2\mathbb{Z} \sqcup \mathbb{S}^1$. So the full list is $$\mathbb{R}/3\mathbb{Z}, \quad \mathbb{R} / 2 \mathbb{Z} \sqcup \mathbb{S}^1, \quad \mathbb{S}^1 \sqcup \mathbb{S}^1 \sqcup \mathbb{S}^1 $$ And, by the way, I'm abusing notations by using $\mathbb{R}/n\mathbb{Z}$ as a shorthand notation for the $n$-sheeted covering map $\mathbb{R}/n\mathbb{Z} \to \mathbb{S}^1$ which is induced from the infinite sheeted covering map $\mathbb{R} \to \mathbb{S}^1$ as you showed in the special case $n=3$. But this is a dangerous abuse of notation, because the topological space $\mathbb{R}/n\mathbb{Z}$ is homeomorphic to $S^1$ for any value of the integer $n \ge 1$.