Gambling is a good starting-point for probability. We can treat $\sigma$-field as a structure of events as we need to define the addition and multiplication for numbers. The completeness of the real numbers is suitable for our calculations, and $\sigma$-field plays the same role.
I hope the following gambling example helps you to understand the filtration and conditional expectation.
Assuming that two people, say player A and player B, bet on the results of two coin tosses.
H: head T: tail
At the time $0$, A and B do not know anything about the result except that one of the events in $\Omega=\{HH,HT,TH,TT\}$ will happen. Hence the information at time $0$ that they both know is $\mathcal{F}_0=\{\emptyset,\Omega\}$.
At the time $1$, the coin had been tossed only once; and they know that the events in the $\sigma$-field $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}\supset \mathcal{F}_0 $ could happen.
At the time $2$, the coin had been tossed twice; and they know that the events in the $\sigma$-field $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}\supset \mathcal{F}_1$ could happen which means they know everything about the gambling results.
Please notice the evolution of information characterized by the filtrations $\mathcal{F}_0,\mathcal{F}_1,\mathcal{F}_2.$ With time passing, the unknown world $\Omega$ is divided more finely. It is something like water flows through pipes.
Assuming that they bet on the following results and the coin is fair.
$$X(\omega)=\left\{ \begin{array}{l}
2, \omega=HH,\mbox{means the first tossing is H, and the second tossing is H}\\
1, \omega=HT,\mbox{means the first tossing is H, and the second tossing is T}\\
1, \omega=TH,\mbox{means the first tossing is T, and the second tossing is H} \\
0, \omega=TT,\mbox{means the first tossing is T, and the second tossing is T}\\
\end{array} \right.$$
Then, we have
$$E[X|\mathcal{F}_0](\omega)=1\qquad\text{for every}\ \omega $$
$$E[X|\mathcal{F_2}](\omega)=X(\omega)\qquad\text{for every}\ \omega $$
$$E[X|\{HH,HT\}]=2P(HH|\{HH,HT\})+1P(HT|\{HH,HT\})$$
$$+1P(TH|\{HH,HT\})+0P(TT|\{HH,HT\})=\frac{3}{2}$$
$$E[X|\{TH,TT\}]=2P(HH|\{TH,TT\})+1P(HT|\{TH,TT\})$$
$$+1P(TH|\{TH,TT\})+0P(TT|\{TH,TT\})=\frac{1}{2} $$
$$E[X|\mathcal{F_1}](\omega)=\left\{ \begin{array}{l}
\frac{3}{2}, \omega\in \{HH,HT\}\\
\frac{1}{2}, \omega \in \{TH,TT\}
\end{array} \right.
$$
I hope those would be helpful.
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
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
Best Answer
As Jonas mentioned, allowing arbitrary unions is not "consistent", in the sense that there is no proper definition of probability. This is also related to the fact that infinite sums make much more sense when countable, since it's not clear how to attach a finite number to an uncountable sum of positive reals.
On the other hand, many desirable events are describable using countable unions and intersections. For example, events like "the random walk returns to the origin" is a union of countably many events "the random walk returns to the origin at time $t$", and any one of those is a finite union of "basic" events.
In general, first order properties always correspond to taking countable unions and intersections; this means that if you have a statement of the form "$\forall x \exists y \cdots P(x,y,\ldots)$", where $x,y,\ldots$ are integers, and the $P$s are basic events (e.g. for a random walk, depend on finitely many times), then the corresponding event is guaranteed to be in the $\sigma$-algebra, i.e. is guaranteed to have assigned to it a "probability".