Let $\Sigma$ be the $\sigma$-algebra. Choose $x \in X$, and define $M_x = \cap_{M \in \Sigma, x \in M} M$. Clearly $M_x \neq \emptyset$, and $M_x \in \Sigma$.
Furthermore, the collection $F = \{M_x \} \subset \Sigma$ is a partition of $X$ (and finite, of course). To see this, suppose $M_x \cap M_y \neq \emptyset$. Then we must have $M_x = M_y$, or else either $M_x \setminus M_y $ or $M_y \setminus M_x $ would be strictly smaller sets contradicting the definition of either $M_x$ or $M_y$.
Furthermore, it is clear that if $M \in \Sigma$, then $M = \cup_{x \in M} M_x$, hence every element of $\Sigma$ is the (disjoint) union of members of $F$ (the empty set taken as the union of no members of $F$), hence $|\Sigma| = 2^{|F|}$.