Your idea to prove by contradiction is correct. Here are the details.
Suppose there is a countable family $\{A_n\}_{n\in\mathbb{N}}$ such, for all $n\in\mathbb{N}$, $A_n\in \Sigma$ and
$$ \Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})$$
For each $n\in\mathbb{N}$, define
\begin{align}
&B_n = A_n & \textrm{ if $A_n$ countable};
\\& B_n = A_n^c & \textrm{ if $A_n$ cocountable}
\end{align}
Then we have, for all $n\in\mathbb{N}$, $B_n$ is countable and, it is easy to see that:
$$ \Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})= \sigma (\{B_n \:|\: n\in\mathbb{N}\}) \tag{1}$$
Let $C=\bigcup_{n\in\mathbb{N}}B_n$. Since $C$ is a countable union of countable sets, we have that $C$ is countable.
Since, for each $n\in\mathbb{N}$, $B_n$ is a countable subset of $C$, we have $B_n\in \sigma(\{\{p\} \:|\: p\in C\})$ and so we have
$$\sigma (\{B_n \:|\: n\in\mathbb{N}\})\subseteq \sigma(\{\{p\} \:|\: p\in C\}) $$
On the other hand, for each $p\in C$, $\{p\}\in \Sigma$ (because $\{p\}$ is obviously countable). So, considering $(1)$, for each $p\in C$, $\{p\}\in \sigma (\{B_n \:|\: n\in\mathbb{N}\})$, and we can conclude that
$$\sigma(\{\{p\} \:|\: p\in C\}) \subseteq \sigma (\{B_n \:|\: n\in\mathbb{N}\})$$
and so we have
$$\Sigma= \sigma (\{B_n \:|\: n\in\mathbb{N}\})= \sigma(\{\{p\} \:|\: p\in C\}) $$
Let $\Sigma_0= \{E \:|\: E\subset C\} \cup \{E\cup C^c \:|\: E\subset C \}$.
It is easy to prove that $\Sigma_0$ is a $\sigma$-algebra, and for each $p\in C$, $\{p\}\in \Sigma_0$. So
$$\Sigma= \sigma(\{\{p\} \:|\: p\in C\}) \subseteq \Sigma_0 \tag{2}$$
Now, note that, since $C$ is countable, $\mathbb{R}\setminus C\neq \emptyset$, that is, $C^c \neq \emptyset$. Let $q$ be any element in $C^c$. We have $\{q\}\in \Sigma$ (because $\{q\}$ is obviously countable) but $\{q\}\notin \Sigma_0$. Contradiction.
Remark 1: We can easily prove that $$\sigma(\{\{p\} \:|\: p\in C\}) = \Sigma_0$$
but all we need is the inclusion presented in $(2)$.
Remark 2: All we used from $\mathbb{R}$ is that it is uncountable. The proof above works for any uncountable space $\Omega$.
Best Answer
$$\mathcal F =\bigcup_{n\ge 0} \mathcal F_n$$ isn’t closed under countable union.
For example, all the singletons $A_n = \{n^2\}$ belong to $\mathcal F$. However, the countable union $\cup_n A_n = \{n^2: n \in \mathbb N\}$ does not belong to $\mathcal F$ as it belongs to none of the $\mathcal F_n$.
$$\mathcal F \neq \sigma(\{0\}, \{1\}, \dots )$$
Note: you don't want to use the singletons $A_n = \{n\}$ to create your counterexample because $\cup_n A_n = \mathbb N$ belongs to all of the $\mathcal F_n$ (a sigma algebra over a set $X$ always contains the empty set and $X$)