Simple problem, searchig for a solution. Let $A$ be a proper, closed and path connected subset of the $S^1$, then $A \cong [0,1]$.
So currently I am trying too use the fact that $f:[0,1] \rightarrow S^1$ by $f(t)=(cos(2\pi t),sin(2\pi t) )$ is a continuous surjective map. However, this leads nowhere and I have become stuck.
I am looking for an elementary proof. Thanks
This only holds if $A$ is a proper and non-singleton subset of $S^1$. Note that $A$ being closed in $S^1$ is compact. Pick $p \in S^1\setminus A$ and note that $S^1 \setminus \{p\} \simeq \Bbb R$ (stereographic projection) and a (path-)connected compact subset of $\Bbb R$ is a singleton or of the form $[a,b], a < b$ and thus homeomorphic to $[0,1]$.