If the universal cover of $X$ is contractible, then $\pi_n(X) = {0}$ [duplicate]
Solution 1:
You're right, the claim made by wikipedia is valid for any covering map. And the proof is also valid.
Let $p : \widetilde X \to X$ a covering map.
As you said, the lifting criterion clearly implies that any map $f$ from a simply connected space to $X$ lifts to $\widetilde X$. So, for example, any $f : S^n \to X$ ($n \geq 2$) lifts to $\widetilde f : S^n \to \widetilde X$.
In your case, $X$ is contractible, so there is an homotopy $(\widetilde f_t)$ from $\widetilde f = \widetilde f_0$ to a constant map $\widetilde f_1$. The homotopy $f_t = p \circ \widetilde f_t$ then goes from $f_0 = f$ to a constant map. We have proved that $\pi_n X = 0$ as soon as $\pi_n \widetilde X = 0$.
In fact, one can see along the same lines that every covering map $p : \widetilde X \to X$ induces isomorphisms $p_* : \pi_n \widetilde X \to \pi_n X$ for all $n \geq 2$ (doesn't work for $n=1$, of course: $S^1$ isn't simply connected).