Let $G \leq GL(V)$ be an affine algebraic group, over an algebraically closed field.
Say that $M$ is a proper subgroup of $G$ which is maximal among the closed proper subgroups of $G$. Does $M$ have to be a maximal subgroup? Ie. is it possible to find a subgroup $M \leq N \leq G$ such that $N \neq M$ and $N \neq G$? What if you assume eg. $G$ simple?
On
For $G$ simple, take $G = \operatorname{PGL}_2(\mathbb{C})$ and let $M < G$ be a maximal closed subgroup that is finite. Such $M$ exist by the classification of finite subgroups of $\operatorname{PGL}_2(\mathbb{C})$, and $M \cong \operatorname{Sym}(4)$ or $M \cong \operatorname{Alt}(5)$. Then $M$ is maximal closed but not maximal, because $M < \operatorname{PGL}_2(\mathbb{Q}(\alpha_1, \cdots, \alpha_t))$ for some $\alpha_i \in \mathbb{C}$.
Let $G$ consists of upper triangular $2\times2$ matrices with ones on the diagonal. As $G$ is 1-dimensional the trivial subgroup $M$ is the only Zariski closed subgroup. Yet any additive subgroup of the base field gives rise to an intermediate subgroup (assume characteristic zero to avoid the possibility of finite subgroups that would surely be also Zariski closed).
Don't know whether $G$ being simple changes this.