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Oops, normalising a maximal torus in $G$, not just in $H$
LSpice
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Let $T_H$ be a maximal torus in $H$.

Suppose that $H^\circ$ is not a torus. Then $T_H$ has non-trivial roots on $\operatorname{Lie}(H)$, and the subgroup of $H$ generated by opposite root groups is $\operatorname{SL}_2$.

Thus, if $H$ does not contain $\operatorname{SL}_2$, then its identity component equals $T_H$; and you have indicated that that torus is regular in $G = \operatorname{GL}_2$, which I take to mean that its centraliser $T_G \mathrel{:=} C_G(T_H)$ in $G$ is a maximal torus in $G$. Thus, $H$ normalises the maximal torus $T_G = C_G(H^\circ)$ in $G$.

LSpice
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