The complement of an at most countable set of reals is uncountable

$\mathbb{R}\setminus C$ is nonempty. If it were empty then $\mathbb{R}\subseteq C$ by [F3], so $C=\mathbb{R}$ and $\mathbb{R}$ would be at most countable, contradicting [F2].

Suppose, for contradiction, that $\mathbb{R}\setminus C$ were at most countable. If $C=\mathrm{\varnothing }$ then $\mathbb{R}=\mathbb{R}\setminus C$ is at most countable, contradicting [F2]; so $C\mathrm{\ne }\mathrm{\varnothing }$. By step 1.1 the complement is nonempty too, so [F1] gives surjections $f:\mathbb{N}\to C$ and $g:\mathbb{N}\to \mathbb{R}\setminus C$. Define $t:\mathbb{N}\times \mathbb{N}\to \mathbb{R}$ by $t(0,k):=f(k)$ and $t(\sigma (i),k):=g(k)$. Every real lies in $C$ or in $\mathbb{R}\setminus C$ by [F3], and such a real is $f(k)$ or $g(k)$, hence a value of $t$; so $t$ is onto. Composing with the bijection $\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ of [F4] gives a surjection $\mathbb{N}\to \mathbb{R}$, contradicting [F2]. Hence $\mathbb{R}\setminus C$ is not at most countable, that is, it is uncountable.

The special cases follow: if $C$ is finite then it is at most countable and $\mathbb{R}\setminus C$ is uncountable; if $C$ is the set of embedded rationals then $\mathbb{R}\setminus C$ is the set of irrationals, uncountable, and this agrees with The irrationals are uncountable; the deleted set may be dense, since the embedded rationals are dense by [F5], yet its complement is uncountable.

Therefore the complement of an at most countable subset of $\mathbb{R}$ is uncountable. The argument instantiates at most two surjections existentially and uses the criterion A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$, whose own proof is choice-free, so no choice principle is used; this is why the result does not need the Axiom of Countable Choice that the countable-union theorem of this page assumes for general countable families.