Note that if $A_n$ is any family of sets, then
$$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=0}^{n-1} A_k\bigr)$$
where the summands on the right-hand side are disjoint, and each of them is constructed from finitely many of the $A_i$s by a sequence of complements and finite unions.
So if you choose you can restrict yourself to requiring countable unions of disjoint set, if you also require that the algebra is closed under finite unions of arbitrary sets.
A Dynkin system that is not a $\sigma$-algebra:
If you don't require arbitrary finite unions, what you get is not necessarily a $\sigma$-algebra. Consider for example the system of subsets of $\mathbb R$ consisting of $\varnothing$, $\{0,x\}$ for every $x\ne 0$, and the complements of these sets. It is closed under your proposed axioms, because the only nontrivial disjoint union there is to take is $\{0,x\} \cup \{0,x\}^\complement = \mathbb R$.
A small finite example is
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
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.
Best Answer
HINT: Given $\{A_n\mid n\in\Bbb N\}$ in your algebra, consider $B_n=\bigcup_{k<n}A_k$.