Let $f: D^2 \rightarrow X$ be a covering map. I am trying to show that $f$ must in fact be a homeomorphism. To do so, I believe it suffices to show that $f$ is injective. Moreover, if only one point of $X$ has a finite pre-image, we can use a connectedness argument to show that $f$ is injective on all of $D^2$. So far, I have attempted to prove this using compactness to obtain finitely many open sets of $X$ which cover $X$ and are each evenly covered under $f$, but have been unsuccessful. Any suggestions?
Also, I am wondering how this generalizes to other compact, simply connected spaces. Is the same true if we replace $D^2$ by $D^n$, $n=1,3,4,5...$?