I know that for this problem I have to use contradiction. Could anyone check my work and guide me through the problem if it's wrong? So far, this is what I have.Thanks!!
Contradiction: $\mathcal B\subseteq \mathcal A$, then $\mathcal B$ is not a family of pairwise disjoint sets.
If $\mathcal B$ is not pairwise disjoint then $\mathcal A \neq \mathcal B$ or $\mathcal A \cap \mathcal B\neq \varnothing $
then, x $\in \mathcal A$ and x$\notin \mathcal B$
However, x $\in \mathcal B\subseteq \mathcal A$ , which is a contradiction since we said that x$\notin \mathcal B$.