proof of "conjugacy theorem of BSA" following Humphreys

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Please follow the book here

Section: $16.4$ Proof of the theorem: (1) When $B\cap B'\neq 0$
case (ii) $B\cap B'$ has no nonzero nilpotent elements. Let $T=B\cap B'$. Then the author shows that we may assume $T\subset H$.

My question: $B\cap B'\subset B$ and has no nonzero nilpotent elements in it and $B$ is standard with respect to CSA, $H$. And we know that a BSA contains $x_s$ and $x_n$, the semisimple and the nilpotent part of the elements $x$ in it. So $T\cap N=0$. Hence $T\subset H$.

Is there anything I've missed? I'm not getting why did he prove that "we may assume" ?