An ultrafilter containing all cocountable subsets

Set $\mathcal{F}=\{A\subseteq X:X\setminus A\text{ is countable}\}$. It contains $X$ because the empty set is countable, and omits $\mathrm{\varnothing }$ because $X$ is uncountable. If $A\in \mathcal{F}$ and $A\subseteq B\subseteq X$, then $X\setminus B\subseteq X\setminus A$, so F4 gives $B\in \mathcal{F}$.

If $A,B\in \mathcal{F}$, then $X\setminus (A\cap B)=(X\setminus A)\cup (X\setminus B)$ is countable: apply F3 to the sequence with these first two terms and empty remaining terms, using the countable choice supplied by A1. Therefore $A\cap B\in \mathcal{F}$. Together with step 1.1 this verifies every filter axiom in F1.

Apply F2 to this proper filter, using A1, to obtain an ultrafilter $\mathcal{U}\supseteq \mathcal{F}$. If countable $D\subseteq X$ belonged to $\mathcal{U}$, its complement would belong to $\mathcal{F}\subseteq \mathcal{U}$; F1 would put $D\cap (X\setminus D)=\mathrm{\varnothing }$ in $\mathcal{U}$, contradicting properness. Thus every member of $\mathcal{U}$ is uncountable. Only pairwise intersections were used.