Start by deformation-retracting $G$ to its maximal compact, so that $G/T$ will be a flag manifold with a Bruhat decomposition, obtainable by Morse theory as in Agol's answer.

If you want the $\pi_2$ anyway, consider the long exact sequence on homotopy of $T \to G \to G/T$, giving $$ \ldots \to \pi_2(T) \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T) \to \pi_1(G) \to \ldots $$ The first obviously vanishes, and the last does too by assumption. We know $\pi_2(G/T)$ is free of dimension $rank(G)$ by the Bruhat decomposition, so the surjection $\pi_2(G/T) \to \pi_1(T)$ is between free abelian groups of the same finite rank. Hence it's also injective. So its kernel $\pi_2(G)$ is zero.

Start by deformation-retracting $G$ to its maximal compact, so that $G/T$ will be a flag manifold with a Bruhat decomposition, obtainable by Morse theory as in Agol's answer.

If what you want anyway is the $\pi_2$, consider the long exact sequence on homotopy of $T \to G \to G/T$, giving $$ \ldots \to \pi_2(T) \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T) \to \pi_1(G) \to \ldots $$ The first obviously vanishes, and the last does too by assumption. We know $\pi_2(G/T)$ is free of dimension $\leq rank(G)$ by the Bruhat decomposition, so the surjection $\pi_2(G/T) \to \pi_1(T)$ is between free abelian groups of the same finite rank. Hence it's also injective. So its kernel $\pi_2(G)$ is zero.