proving that a covering map with certain domain and range is homeomorphism

The facts you need are:

1. For every covering map $p\colon E \to B$ and point $e \in E$ the map $p^\ast\colon\pi_1(E, e) \to \pi_1(B, p(e))$ induced on fundamental groups is injective.
2. If $p\colon E \to B$ is a covering map and $E$ is simply connected then each fiber of $p$ has the same cardinality as the fundamental group of $B$.
3. Every injective covering map is an isomorphism.

You are almost done. Note that since $p(f)$ is nullhomotopic, the class $[p(f)]$ is in the trivial image $p_*(\pi_1(E,e_0))$. The elements in the image of $p_*$ are classes of loops whose lifts at $e_0$ in $E$ are also loops, i.e. the lift $f$ of $p(f)$ is a loop.

Well, you know that the number of sheets exactly equal index of subgroup $p_*(\pi(E))$ in $\pi(B)$, since $B$ is simply connected, it implies that the index is $1$, so $p$ is one-to-one.