Let $(A_n)_n$ be sequence of subsets of a set $\Omega$. For all $a \in \{0,1 \}^{\mathbb{N}}$, define
\begin{equation} \label{eu_eqn} F(a) := \bigcap^\infty_{i=1}A_i^{(a_i)} , \end{equation} where $A^{(0)}:= A_i$ and $A^{(1)}:= A_i^c$.
I have already shown that the collection $\{F(a): a \in \{0,1 \}^{\mathbb{N}} \}$ is a partition of $\Omega$. I have also already shown that \begin{equation} \sigma((A_n)_n) = \{\bigcup_{a \in J}F(a): J \subset \{0,1 \}^{\mathbb{N}} \}. \end{equation} I have done this by proving that the right hand side is just a $\sigma$-algebra that contains $(A_n)_n$. My goal is now to prove the following statement:
$\sigma((A_n)_n)$ is either finite or uncountable.
Through my construction it is easy to see that $\sigma((A_n)_n)$ is uncountable if it is infinite, since if it is infinite, then $(A_n)_n$ should contain an infinite amount of different elements. In that case, the right hand side has the same cardinality as the set of all sequences in $\{0,1 \}^{\mathbb{N}}$ (which is uncountable many). I know this because for every $a \in \{0,1 \}^{\mathbb{N}}$ I can define $J:= \{a \}$. However, there is one thing that stands in the way of my proof. In order for it to have that cardinality, I have to show that there is a one-to-one correspondence between every sequence $a \in \{0,1\}^{\mathbb{N}}$ and the set $\{F(a): a \in \{0,1 \}^{\mathbb{N}} \}$. So I first have to prove the following statement:
\begin{equation} \forall a,b \in \{0,1 \}^{\mathbb{N}}: a \neq b \Rightarrow F(a) \neq F(b). \end{equation} However, when I try this, I get into all sorts of problems with the empty set. This statement is fairly easy to prove if $F(a)$ or $F(b)$ is not the empty set. However, when I assume $F(a) = \emptyset$, it is not apparent for me to see that $F(b) \neq \emptyset$. In other words, the collection $\{F(a): a \in \{0,1 \}^{\mathbb{N}} \}$ is clearly a partition of $\Omega$, but it is not clear for me that more than one $a \in \{0,1 \}^{\mathbb{N}}$ can correspond to the empty set (maybe even an uncountable amount, such that the right hand side could possibly be countable). Note that my definition of partition of a set $\Omega$ is just a collection of disjoint sets such that the (disjoint) union of those sets is $\Omega$ (so the empty set can be in the partition). I feel like I'm close to proving the statement, but I am just stuck at this silly little nuance. Or maybe I have made an error of some kind somewhere...
Clearly, if $\mathcal{F} := \{F(a): a \in \{0, 1\}^\mathbb{N}\}$ is finite, than so is $\Sigma := \sigma((A_n)),$ so your question just asks you to prove that $\Sigma$ is uncountable when $\mathcal{F}$ is infinite. Consider an arbitrary injection $i : \mathbb{N} \to \mathcal{F}.$ This injection gives rise to a natural injection $\mathfrak{i} : 2^\mathbb{N} \to \Sigma$ defined by $$\mathfrak{i} ((b_n)_{n \in \mathbb{N}}) := \bigcup_{n \in \mathbb{N}} i(n)^{b_n},$$ where $A^0 := \emptyset, \text{ } A^1 := A.$ Indeed, if $(b_n) \neq (b_n'),$ let $m := \min \{n: b_n \neq b_n'\}$ and WLOG, $b_n = 0 = 1 - b_n'.$ Then $i(m) \cap \mathfrak{i} ((b_n')_{n \in \mathbb{N}}) = i(m),$ while $i(m) \cap \mathfrak{i} ((b_n)_{n \in \mathbb{N}}) = \emptyset$ since all of the $i(n)'s$ are disjoint. It follows that $\Sigma$ is uncountable. I hope this helps. :)