Daniel@30: A typical way to prove \$$A = \cup_{p\in P}{A_p} \$$ is by double inclusion. Since \$$A_p\$$ are all subsets of \$$A\$$ one inclusion is clear. For the other, suppose there is an \$$a\in A\$$ and \$$a\not\in\cup_{p\in P}{A_p} \$$, what can be said about \$$\{a\}\$$ ?

Edit: For some reason the brackets are not showing, both "a" are supposed to be the set containing only "a".

SPOILER: \$$\{a\}\$$ is a \$$\sim\$$-connected and \$$\sim\$$-closed subset, contradiction.