In any topos $\mathcal{E}$, any object $d$ has a lattice of subobjects $(\operatorname{Sub}(d), \subset)$.
One can give a good description of the object $a\cap b$ for $a,b$ subobjects of $d$, and similarly $a\cup b$.
This description uses products, so it seems natural, if we want $\operatorname{Sub}(d)$ to be a complete lattice to assume that the topos $\mathcal{E}$ has arbitrary products (although I don't know if it's a necessary condition).
Therefore my questions are :
Is "$\mathcal{E}$ is closed under arbitrary (small) products" a sufficient condition for "for all $\mathcal{E}$-object $d$, $\operatorname{Sub}(d)$ is a complete lattice" ?
Are there known and interesting necessary and sufficient conditions so that they are complete lattices ?
Yes, the existence of all products suffices to construct all intersections of subobjects. This is simply because a topos is finitely complete, so the existence of products implies the existence of all limits. Since toposes are finitely cocomplete and a lattice has all intersections whenever it has all unions, we can make the same comment regarding coproducts.
This is certainly not a necessary condition, as the topos of finite sets certainly has complete subobjects lattices without admitting any nontrivial infinite products or coproducts. But this admits a logical embedding into the topos of sets, which does have all products...Perhaps there are better examples that don't come this way, I'm not sure.