Note that $(\mathcal{L}, \subseteq)$ is distributive if and only if the lattice $M_3$ ( en.wikipedia.org/wiki/Distributive_lattice#mediaviewer/… ) cannot be embedded in $(\mathcal{L}, \subseteq)$. So one idea might be to find three group topologies $\tau_1,\tau_2,\tau_3$ with identical infimum and identical supremum, or show that this cannot be done.

Interestingly, for infinite cardinals $\kappa$ the "Hadwiger statement" is true: If the complete graph on $\kappa$ vertices is not a minor of a graph $G$, then there is a $\kappa$-coloring of $G$ (see arxiv.org/pdf/1312.2829.pdf )

It is not always possible, if we manage to construct a map $F:B^* \to A^*$ with the properties $(1)-(3)$ above but which doesn't respect arbitrary intersections (only finite ones). Then there can't be a map $f$ with the desired properties, because $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$ respects arbitrary intersections.

Thanks for pointing me to Fisher's inequality. I assume this argument can be generalised to prove the statement for intersecting linear hypergraphs with infinite number of points and lines (or maybe in that case, an easy set-theoretic argument does the job, will have to think about it).