Be wary not to confuse this post with this one. In this case, $M$ is simply a subset of the power set of $X$.
Let $X$ be a set and $M$ be a nonempty collection of subsets of $X$. Is it true that if $M$ is closed under complements and countable unions of disjoint sets, then $M$ is a $\sigma$-algebra? [Hint: any countable union of sets can be written as a countable union of disjoint sets.]
The hint confuses me because it seems to suggest that the question is true. But I don't think it is. Just choose $X= \{a,b,c,d\}$ and choose
$$M = \{\varnothing, X, \{a,b\}, \{b,c\}, \{c,d\}, \{a,d\}\}$$
Then $M$ is certainly closed under countable disjoint union and complements, but $\{a,b\}\cup \{b,c\} = \{a,b,c\} \not\in M$. So $M$ is not closed under countable unions and so $M$ is not a $\sigma$-algebra. I suspect I'm missing something here because this was too easy to do, especially given the hint.