Here's another proof based on the structure of the flag variety G/T$G/T$ of G$G$. A compact Lie group G$G$ has a maximal torus T$T$, and G$G$ is a principal T$T$-bundle over the quotient G/T$G/T$. Borel showed that G/T$G/T$ is a complex manifold, and gave a CW decomposition of it with no odd-dimensional cells. (This is not deep but still astonishing, and the start of a long story; I like the context given by Hirzebruch's eulogy for Borel, available on page 9 here.)
Since pi_2(T) = 0$\pi_2(T) = 0$, we have an exact sequence
0 --> pi_2(G) --> pi_2(G/T) --> pi_1(T)$$0 \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T)$$
We can conclude immediately that pi_2(G)$\pi_2(G)$ is torsion-free, since pi_2(G/T) = H_2(G/T)$\pi_2(G/T) = H_2(G/T)$ is a free group on the 2-cells in G/T$G/T$. After Allen's answer (Hopf's theorem) this shows pi_2(G) = 0$\pi_2(G) = 0$.
With a little more Lie theory one can show directly that the connecting map pi_2(G/T) --> pi_1(T)$\pi_2(G/T) \to \pi_1(T)$ is injective. pi_1(T)The group $\pi_1(T)$ has a linearly independent subset of simple coroots, and the 2-cells in G/T$G/T$ are indexed by simple roots. The connecting homomorphism matches these up in the natural way, which one can see by considering rank 1 subgroups (subgroups of the form SU(2)$SU(2)$ or PSU(2)$PSU(2)$) of G$G$. As a consequence you get a formula for pi_1(G)$\pi_1(G)$ in terms of roots and coroots.
