Starting from a initial collection of sets being allowed to take countable unions and intersections lets you create many more sets that being allowed to take only finite unions and intersections. Therefore it seems plausible to me that the former can take you out of your starting collection even if the latter does not.
You can find examples in any standard measure theory text. Here is a very simple one. Let
$$A_n=\{1,2,3,\ldots,n\}$$
Let $\Omega$ be the collection of the sets $A_n$ for $n \in \mathbb{N}$.
Then for a finite set of indices, $n_1,n_2,\ldots,n_k$ say, the union
$$A_{n_1} \cup A_{n_2} \cup \cdots \cup A_{n_k} = A_{\max\{n_1,\ldots,n_k\}}$$
and hence belongs to $\Omega$. So $\Omega$ is closed under finite unions.
But the countably infinite union
$$\cup_{n=1}^{\infty} A_n = \{1,2,3,\ldots\}$$
and hence does not belong to $\Omega$. So $\Omega$ is not closed under countable unions.
A formal proof would be something like this. If $A$ and $B$ are countable (and disjoint) sets, then there exist injective functions $f$ and $g$ from $A$ and $B$, respectively, to $\mathbb{N}$. Let $C=A\cup B$ and define a function $h:C\rightarrow\mathbb{N}$ as follows:
$$h(z) = \begin{cases}2f(z) & \text{if } z\in A\\ 2g(z)+1 & \text{if } z\in B\end{cases}$$
Then, if $x\neq y$, we clearly have $h(x)\neq h(y)$ since if $x,y\in A$ this follows from the fact that $f$ is injective (analogously for $x,y\in B$) and if $x\in A$ and $y\in B$, we have that $h(x)$ is even while $h(y)$ is odd (analogously for $x\in B$, $y\in A$). Hence $h$ is an injective function from $C$ to $\mathbb{N}$ and thus, by definition, $C$ is countable.
Best Answer
Let $A, B, C, D, E$ be sets.
$A = \{0, 1, 2, \cdots 9\}$
$B = \{0, 2, 4, \cdots, 18\}$
$C = \{1, 3, 5, \cdots, 19\}$
$D = \{10, 11, 12, \cdots 19\}$
$E = \{0, 5, 10, 15, 20\}$
Let $P = \{A, B, E\}$ and $Q = \{C, D\}$.
Then $P\cup Q = \{A, B, C, D, E\}$.
Here, the elements of $P, Q$ are sets, so their union is a set of sets (a set whose elements are sets). So their union is not a set of all the set elements contained in $A, B, C, D, E$. (I.e., none of the numbers $0$ to $20$ are in $P, \;Q,\;$ OR $P\cup Q$.)
If you want to describe the union of all elements in $A, B, C, D, E$, you would write, e.g., $A \cup B \cup C\cup D\cup E = \{0, 1, 2, 3, \ldots, 18, 19, 20\}$.