Why we impose finite countability to be closed in $\sigma$-algebra $\mathscr{F}$

In $\sigma$-algebra, denoted as $\mathscr{F}$, one of the 3 main conditions that we impose to have a $\sigma$-algebra is the fact that for countable subsets $\{A_i, i\ge 1\}$, we need these subsets to be closed under unions:

$$ \bigcup_i{A_i}\in \mathscr{F} $$

Question: However, $\mathscr{F}$ itself might have uncountable subsets, so I am confused to distinguish between the condition we just imposed for subsets to be closed under unions and the fact that $\mathscr{F}$ might have uncountable subsets? Can you please clarify this?

I think you misunderstood a thing about the $\sigma$-algebra

One condition for $\mathscr{F}$ to be a $\sigma$-algebra is that the union of contable many sets $A_i$ (where $A_i \in \mathscr{F}$ can in fact be uncountable) must be again an element of $\mathscr{F}$

For example $\mathscr{F} = P(\mathbb{R})$ is a $\sigma$ -algebra (in fact any powerset is). And $P(\mathbb{R})$ contains elements that are uncountable.

$\mathscr{F}$ doesn't need to have countably infinite sets. Consider a set $X = \{1,2,3\}$ and $\mathscr{F}= \{ \emptyset, X, \{1\}, \{2,3\} \}$ That's a $\sigma$ -algebra (try to check the conditions).