Let $H$ and $K$ be two proper non-trivial subgroups of the
alternating group $A_n; n\geq 5$.
Then there exists a maximal subgroup $M$ of $A_n$
such that $H\not\leq M$ and $K\not\leq M$.
To see this
let $\Omega:=\{1,2,\cdots,n\}$
and as always ${\rm Supp}_{\Omega}(H):=
\{\omega\in\Omega\mid \omega^{h}\neq \omega,$ for some $h\in H\}$.
If there exists $l\in {\rm Supp}_{\Omega}(H)\cap {\rm Supp}_{\Omega}(K)$,
put $M\!:={\rm Stab}_{A_n}(l)$, otherwise take
$l\in {\rm Supp}_{\Omega}(H)\setminus {\rm Supp}_{\Omega}(K)$

and $t\in {\rm Supp}_{\Omega}(K)\setminus {\rm Supp}_{\Omega}(H)$
and put $M:={\rm Stab}_{A_n}\{l,t\}$. (May be other types of maximal subgroups of $A_n$,
according to O'Nan-Scott Therorem, satisfy this property but I do not know).

By a GAP code I could check that this property is also true for a few finite non-abelian simple groups apart from the alternating groups, like ${\rm PSL}(2,7), {\rm PSL}(2,8), {\rm M}_{11},{\rm M}_{12}$.

Could we generalize this property for other types of finite non-abelian

simple groups?

Thank you very much!