From Wikipedia (about homotopy of spheres):
$\pi_2 (S^1) = 0$: This is because $S^1$ has the real line as its universal cover which is contractible (it has the homotopy type of a point). In addition, because $S^2$ is simply connected, by the lifting criterion, any map from $S^2$ to $S^1$ can be lifted to a map into the real line and the nullhomotopy descends to the downstairs space.
Question 1: Am I right in assuming that I don't need a universal cover? (just any covering space is good enough to apply the lifting criterion)
The "lifting criterion" tells me that if $p: (\tilde{X},a) \to (X,b)$ is a covering map and $f: (Y,c) \to (X,b)$ is any continuous map then $f$ lifts to $\tilde{f}: (Y,c) \to (\tilde{X}, a)$ if and only if $f^\ast \pi_1 (Y) \subset p^\ast \pi_1 (\tilde{X})$.
Hence For $\tilde{X} = \mathbb R, X = S^1, Y = S^2$, $p^\ast \pi_1 (\tilde{X}) = \{0\}$ and hence since $f : S^2 \to S^1$ lifts to $\tilde{f} : S^2 \to \mathbb R$, $f^\ast \pi_1 (Y)$ must be the trivial group, too.
Question 2: But how do I get from there to $\pi_{\color{red}{2}} = 0$? The lifting criterion only tells me something about $\pi_1$.
Thanks for your help.