It is easy to see that
$p:\mathbb{R} \rightarrow S^1: p(t)=(\cos t, \sin t)$ is a covering map of $S^1$
(Indeed take a point $x_0$ in $S^1$, take $U=S^1\setminus \{-x_0\}$ as an open neighborhood and then $p^{-1}(U)=\cup_{n \in Z} J_n$ where $J_n=\{t\in\mathbb{R} : t_0-n-1/2
I can't see why we can't take a similar $p:(a,b) \to S^1$. For example, Massey states that if we take the open interval $(0,10)$ onto the circle, then some points in $S^1$ fail to have an elementary neighborhood.