A subgroup that is both unipotent and diagonalizable is trivial

Write $O(H)=k[M]$. For each $m\in M$, its character representation $k_m$ is one-dimensional and nonzero. Unipotence gives a nonzero fixed vector in $k_m$ by [F1], so its character is trivial: $e_m=e_0$ as a function on the group scheme, with equality on every base algebra. Since the elements $e_m$ form a basis of $k[M]$, this equality forces $m=0$. Thus $M=0$, $O(H)=k$, and $H=1$. This tests individual character lines and uses neither arbitrary character-line decompositions nor a faithful-representation existence theorem.

Assume AC for this geometric corollary. If $H$ is a unipotent closed subgroup of an arbitrary torus $T$, pass to a field extension splitting $T$. Then $H$ remains unipotent and is diagonalizable by [F3], hence is trivial by step 1.1. Faithfully flat descent gives $H=1$ over $k$. The intersection of a unipotent subgroup with a torus is a closed unipotent subgroup of that torus, so the same reasoning makes the intersection trivial. This includes nonreduced subgroup schemes and nonsplit tori.