This question is a natural follow up to Constructing a subset of an uncountable set which is neither countable nor co-countable.
Let $\Omega$ be an uncountable set. Using the axiom of choice, how can we construct a subset $S\subset \Omega$ which is neither countable nor co-countable?
If $f: \mathcal{P}(\Omega)-\emptyset \to \Omega$ is a choice function, construct two sequences $(a_\xi)$, $(b_\xi)$ by transfinite recursion with
$a_\xi=f(\Omega-(\{a_\eta:\eta < \xi\} \cup \{b_\eta:\eta < \xi\}))$ and $b_\xi=f(\Omega-(\{a_\eta:\eta \leq \xi\} \cup \{b_\eta:\eta < \xi\}))$.
The process does not stop at any countable ordinal by uncountability of $\Omega$.
well-order your set $\Omega$, and pick all those points whose index is a limit ordinal. You can easily show that if $\alpha$ is an uncountable ordinal, then the set of limit ordinals is neither countable, nor co-countable.
This allows you even slightly better control on the cardinality of the set you create (or the cardinality of its complement).
