Suppose that $H^\circ$ is not a torus, and let $T_H$ be a maximal torus in $H$. 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 is a torus, and $H$ is contained in the normaliser of that torus.

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 $G$.

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

Thus, if $H$ does not contain $\operatorname{SL}_2$, then its identity component is a torus, and $H$ is contained in the normaliser of that torus.

Suppose that $H^\circ$ is not a torus, and let $T_H$ be a maximal torus in $H$. 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 is a torus, and $H$ is contained in the normaliser of that torus.