If A, B \in \mathcal {G} then A \cap B \in \mathcal{G} as \mathcal{G} is closed under finite intersections and therefore A \cup B \in \mathcal {M} using
the DeMorgan's laws and property II of \mathcal {M} as you have pointed out.
I could not prove that if A, B, C, D, E\in \mathcal...