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 $\{\emptyset\}$.
Best Answer
The word 'countable' is the same as 'in bijection with the natural numbers' or 'in bijection with the integers.' There are infinitely many integers, so it's "bigger" than finite. But it's also somehow the smallest infinity.
A common case where this might come up is with respect to open and closed sets. A finite union of closed sets is closed. But an infinite union of closed sets might not be closed. For example, if we consider the sets $I_n = [\frac{1}{n}, 1 - \frac{1}{n}]$, then each $I_n$ is closed. But $\cup_{n \in \mathbb{N}} I_n = (0,1)$, an open set.
With respect to your De Morgan's law question: It is a fundamental fact that $A = B \iff A^c = B^c$, and that $(A^c)^c = A$. So they complemented your De Morgan's law to get that statement.
Finally - algebras and sigma algebras are collections of sets. To be closed under finite intersections means that taking any number of finite intersections of elements of the algebra yields an element (another set) that is in the algebra. But maybe this isn't true for an infinite intersection, etc.