Question: How to show that a covering map which is also a homotopy equivalence is a homeomorphism?

Answer: 
Let $p:\tilde X \to X$ be the covering map. As a homotopy equivalence it induces an isomorphism $\pi\_1(\tilde X,\tilde x\_0)\to\pi\_1(X,x\_0)$ for any $x\_0\in X.$ Now, assume $p(y\_0)=p(y\_1)=x\_0.$ We want to show $y\_0=y\_1.$ Since a homotopy equivalence induces a bijection between the path-components of $\tilde X$ and the path-components of $X$, the points $y\_0,y\_1$ must lie in the same path-component. So let $\alpha$ be a path from $y\_0$ to $y\_1.$ Its image is a loop $p\alpha$at $x\_0.$ By the induced isomorphism $p\_\*$ there is a unique class $[\beta],$ such that $p\_\*[\beta]=[p\alpha],$ i.e. $p\beta$ is homotopic to $p\alpha,$ and this homotopy lifts to a unique path-homotopy between the loops $\beta$ and the unique lift of $p\alpha,$ which is $\alpha.$ This implies that $\alpha$ is a loop and $y\_0=y\_1.$  

Now injectivity makes $p$ a homeomorphism.

The length of the proof depends on how much you know about covering maps. If you know that each loop $\gamma$ at $x\_0$ such that $[\gamma]\in p\_\*(\pi\_1(\tilde X))$ has a lift which is a loop, then it follows immediately that the $\alpha$ in the proof is a loop.
We can modify a little Stefan's reasoning. After reducing problem to path component we can firstly note that following lemma holds:

> 
> **(Conjugacy Theorem).** Let $p:\tilde X \to X$ be a covering map. For any $q \in X$, as $\tilde q$ varies over the fiber $p^{-1}(q)$, the set of subgroups $p\_\*\pi\_1(\tilde X, \tilde q) \subset \pi\_1(X,q)$ is exactly one conjugacy class.
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> 
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but homotopy equivalence induces isomorphism of fundamental groups so the conjugacy class must contain one element and so our covering is $1$-sheeted so is injective and hence homeomorphism.