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.