Condition on the cardinality of $\Omega$ for a countably generated sigma-algebra

I am trying to work out the following question. Any hints are very much appreciated.

Let $\mathcal{F}$ consist of all $A \subseteq \Omega$ such that either $A$ is a countable set or $A^{\complement}$ is a countable set

1. Verify that $\mathcal{F}$ is a $\sigma$-algebra.
2. Show that $\mathcal{F}$ is countably generated if and only if $\Omega$ is a countable set.

My attempt is as follows (please point out if there are any inconsistencies in arguments).

Proof: $\mathcal{F}$ is a $\sigma$-algebra

1. $\varnothing$ is a countable set. Hence $\varnothing\in \mathcal{F}$. Since $\Omega^{\complement} = \varnothing$, hence $\Omega \in \mathcal{F}$

2. For any set $A \in \mathcal{F}$, $A^{\complement} \in \mathcal{F}$ (this is by definition of $\mathcal{F}$, since either $A$ or $A^{\complement}$ is countable)

3. Consider any countable collection of sets $A_1, A_2, A_3, \cdots \in \mathcal{F}$. If $A_i$ 's are all countable then $\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$ (since countable union of countable sets is countable). If for any $j$, $A_j$ is such that $A_j^{\complement}$ is countable then consider the set $B = \bigcap_{i=1}^{\infty} A_i^{\complement}$. Since $A_j^{\complement}$ is countable hence $B$ is atmost countable. Therefore $B \in \mathcal{F}$ and so $B^{\complement} \in \mathcal{F}$. Which implies $\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$. Hence $\mathcal{F}$ is closed under countable unions.

Proof: $\mathcal{F}$ is countably generated if and only if $\Omega$ is a countable set

If $\Omega$ is a countable set i.e. $\Omega = \{\omega_1, \omega_2, \omega_3, \cdots \}$, then as suggested by the above construction $\mathcal{F} = 2^{\Omega}$. And since $2^{\Omega}$ is the $\sigma$-algebra generated by the collection of singletons $\{\omega_i\}$, which is a countable collection, hence $\mathcal{F}$ is countably generated.

However I am unable to prove the only-if statement. Thanks for your suggestions.

Let it be that $$\mathcal A$$ is a countable collection of subsets of $$\Omega$$ with $$\sigma(\mathcal A)=\mathcal F$$.

Then $$\mathcal A\subseteq\mathcal F$$ and because every $$A\in\mathcal A$$ may be replaced by its complement without loosing the property $$\sigma(\mathcal A)=\mathcal F$$ we can take every $$A\in\mathcal A$$ to be countable.

Consequently $$\bigcup\mathcal A$$ is countable.

Now define: $$\mathcal F_0=\{A\in\wp(\Omega)\mid A\subseteq\bigcup\mathcal A\text{ or } A^{\complement}\subseteq\bigcup\mathcal A\}$$.

It is straightforward to prove that $$\mathcal F_0$$ is a $$\sigma$$-algebra.

This with $$\mathcal A\subseteq \mathcal F_0$$ and consequently $$\mathcal F\subseteq\mathcal F_0$$.

Now assume that $$\Omega$$ is not countable.

Then some $$\omega\in\Omega$$ must exist with $$\omega\notin\bigcup\mathcal A$$.

Next to that $$\{\omega\}^{\complement}$$ is uncountable hence cannot be a subset of $$\bigcup\mathcal A$$.

Then $$\{\omega\}\notin\mathcal F_0$$ contradicting that $$\{\omega\}\in\mathcal F\subseteq\mathcal F_0$$.

This contradiction tells us that the assumption that $$\Omega$$ is uncountable is wrong.