In general, "closed under operation * on the set $\Sigma$" means that, if you apply operation * (in this case, taking countable intersections) to elements of the set $\Sigma$ (in this case, the set is the $\sigma$-algebra of subsets of a given set $X$), then the result is still an element of $\Sigma$.
Another example which may be more familiar to you: a vector space is closed under addition (addition of vectors is still a vector)
More precisely, in this case, if $\Sigma$ is a $\sigma$-algebra of subsets of $X$, then if $\{A_i\}_{i=1}^\infty \subset \Sigma$, i.e. $A_1, A_2,\dots, A_n, \dots \in \Sigma$ then $\bigcap_{i=1}^\infty A_i = A_1 \cap A_2 \cap \dots \cap A_n \cap \dots \in \Sigma$.
As I said in a comment above, by allowing countable intersections you get finite intersections too. If $A_1,\dots,A_n \in \Sigma$, then because $X\in \Sigma$ too by definition of $\sigma$-algebra, then using the fact $\Sigma$ is closed under countable intersections, $A_1\cap \dots \cap A_n = A_1 \cap \dots \cap A_n \cap X \cap X \cap X \cap \dots \in \Sigma$.
To answer to your addendum:
There seems to be a bit of confusion. If $X=\{x,y,z\}$, then $\mathcal{P}(X)=\{\emptyset,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$ is the power set of $X$, and a $\sigma$-algebra is a subset of the power set with some additional properties.
So, if $\Sigma=\{\emptyset,\{x\},\{y,z\},\{x,y,z\}\} \subset \mathcal{P}(X)$, then it is a $\sigma$-algebra of subsets of $X$.
To check item 2, what "closed under complementation" means is that, whenever $A\in \Sigma$, we have $A^c=X\setminus A \in \Sigma$. So, we check that for all four elements of $\Sigma$:
$\emptyset \in \Sigma \Rightarrow \{x,y,z\} \in \Sigma$.
$\{x\} \in \Sigma \Rightarrow \{y,z\} \in \Sigma$.
$\{y,z\} \in \Sigma \Rightarrow \{x\} \in \Sigma$.
$\{x,y,z\} \in \Sigma \Rightarrow \emptyset \in \Sigma$.
All these sentences are true, so item 2 is checked.
Also, as a side note, be careful, do not confuse $\emptyset$ with $\{\emptyset\}$.
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
For a small example:
$\mathcal{F}=\{\{1\},\{2\},\{1,2\}\}$ is closed under unions... (Each of $\{1\}\cup\{1\},\{1\}\cup\{2\},\{1\}\cup\{1,2\},\dots$ etc... are elements of $\mathcal{F}$)
...however it is not closed under intersections. ($\{1\}\cap\{2\}=\emptyset\notin\mathcal{F}$)