In order to turn any sequence of sets into a monotone increasing or decreasing sequence of sets, you have to use the finite unions/finite intersections (respectively) and set complements.
A set algebra is precisely a collection of sets which is closed under finite unions, finite intersections, and set complements.
Note also that being closed under finite unions and set complements implies being closed under finite intersections, and being closed under finite intersections and set complements implies being closed under finite unions, by DeMorgan's Laws.
I'm trying to find a source for this, and I can't right now, but I believe if you have a countable union of sets, not necessarily increasing, i.e. $$B = \bigcup\limits_{i=1}^{\infty} B_i$$ then you can always write it as the limit of an increasing union of sets by setting: $$A_1=B_1, A_2= B_1 \cup B_2, \dots, A_n = B_1 \cup \dots \cup B_n $$ Then clearly $\bigcup_{i=0}^{\infty} B_i=\bigcup_{i=1}^{\infty} A_i$. Since each of the $A_i$ is defined using only a finite union, it is guaranteed to be in the system of sets if that system is a set algebra. Then because it is a monotone class, it contains the union of the $A_i$, thus the union of the $B_i$ which is equal, and therefore it is even a $\sigma-$algebra, since $B_i$ was an arbitrary countable family.
If you want to write in the form $\lim_{i \to \infty} C_i = \emptyset$, then it suffices to write $$C_n = A_n \setminus A_{n-1} $$ and we are only guaranteed that each $C_i$ is in our set system if it is a set algebra because otherwise the set system may not be closed under set complements.
And of course $$\bigcup_{i=0}^{\infty} B_i=\bigcup_{i=1}^{\infty} A_i = \bigcup_{i=1}^{\infty} C_i$$
The procedure for showing closure under countable intersections is equivalent; just use the fact that your set system is closed under set complements combined with DeMorgan's Laws.
No. If $M$ is too small (i.e., empty) or too large (i.e., not a set) then it is not a $\sigma$-algebra. But those are trivial cases. Here is a less pathological counterexample:
Let $X=\{1,2,3,4\}$ and
$$M=\big\{\emptyset,\;\;\{1,2\},\;\;\{2,3\},\;\;\{3,4\},\;\;\{1,4\},\;\;\{1,2,3,4\}\big\}$$
Then $M$ is a monotone class closed under complements and finite disjoint unions, but not under intersections.
Best Answer
Following the hint by Harald Hanche-Olsen: take the collection of all unbounded intervals of $\mathbb R$. This includes $\mathbb R$, so we should throw in the empty set too, to keep the class closed under complements. The monotone class is not hard to verify. And finite intersections can create bounded intervals, not in the class.