1. Each \$$A_p\$$ is nonempty because it is \$$(\sim)\$$-connected, and a \$$(\sim)\$$-connected subset is nonempty by definition.

2. If \$$A_p\$$ and \$$A_q\$$ had a nonempty intersection, they would each contain some element \$$x \in X\$$. But then, since they are \$$(\sim)\$$-closed and \$$(\sim)\$$-connected, \$$A_p\$$ and \$$A_q\$$ would each contain *all and only* elements of X equivalent to x, contradicting that \$$A_p\$$ and \$$A_q\$$ are unequal. So \$$A_p\$$ and \$$A_q\$$ can’t have a nonempty intersection.

3. We prove that A and the specified union are subsets of each other: First, every element in the union is obviously in A, because the union is a union of *subsets* of A. On the other hand, if \$$x \in A\$$, then the set \$$A_r\$$ of all elements \$$y \in A\$$ with y \$$\sim\$$ x is \$$(\sim)\$$-closed and \$$(\sim)\$$-connected (by transitivity and reflexivity of equivalence). So \$$A_r\$$ is by definition one of the subsets in the specified union, and thus x (which is in \$$A_r\$$ by reflexivity) is in that union.