Lemma 2

Let $x$ be an element of T and S a subtorus of T. If the identity component of $Z(x)\cap Z(S)$ reduces to T, there exists an element $s$ of S such that $xs$ is regular.

For all $\alpha \in R(G,T)$, let $S_{\alpha}$ be the submanifold of S consisting of the elements $s$ of S such that $(xs)^{\alpha}= 1$. If there is no element $s$ of S such that $xs$ is regular, S is the union of the submanifolds $S_{\alpha}$, hence is equal to one of them. Hence there exists $\alpha$ in $R(G,T)$ such that $(xs)^{\alpha}= 1$ for all $s\in S$; but this implies that $x^{\alpha}= 1$ and $\alpha |S = 1$, so $Z(x)\cap Z(S)\supset$ Z(Ker $\alpha$ ), hence the lemma.