I'm genuinely confused on what this question is asking. Is this saying $F = {A \cup A^c} $ but $A\cup A^c = \Omega$?
Well I attempted the proof anyways and this is my attempt. I know there was another answer but his/her answer just doesn't make sense to me.
$\textbf {(1)}$ Suppose $A = \left\{\emptyset\right\} \rightarrow A^c \neq \left\{\emptyset\right\}$ similarly $A^c = \left\{\emptyset\right\} \rightarrow A \neq \left\{\emptyset\right\}$. Thus $\mathscr{F} \neq \left\{\emptyset\right\}$.
$\textbf{(2)}$ Problems with this one since I don't understand what I'm trying to prove. Prove that it's closed under countable unions.
$\textbf{(3)}$ Suppose $\mathscr{F} = \big\{A^c: A^c \in R$ or $(A^c)^c \in R\big\} \leftrightarrow \mathscr{F} = \big\{A: A \in R$ or $A^c \in R\big\}$ so $\mathscr{R}$ is closed under complementation.
My question is do I go through and prove each step assuming A is countable, then doing it again assuming A is not countable?
I think your question only make sense regarding the 3) property (closure under complement).
You take $A \in \mathcal{F}$, and check if its complement $A^C \in \mathcal{F}$. Which is pretty straight-forward, you need to show that either $A^C$ or $(A^C)^C = A$ is a countable set.
The first property : you need to check if $\emptyset \in \mathcal{F}$. Hint: check if it is countable.
Closure under countable union will not work with your suggestion. You pick a countable number of sets, all in $\mathcal{F}$. And you just do not know if all countable or all complements are countable. You can't assume a thing. So, you need to handle a typical case, where some are countable and some are not. Hint: $(A_1 \cup A_2)^C \subseteq A_1^C$.
And there is one edge case. What if all picked sets are in the end countable?