Not sure where my argument breaks down: Let $A \subset \mathbb{R}$ be a countable set. Prove that $\mathbb{R} \setminus A$ is uncountable.

Question

$\textit{Proof.}$ Since A is countable, we know it has the same cardinality as $\mathbb{N}$, or $\mid A \mid = \mid \mathbb{N} \mid$. Additionally, since $\mathbb{N} \subset \mathbb{R}$, it is enough to prove that $\mathbb{R} \setminus \mathbb{N}$ has the same cardinality as $\mathbb{R}$, an uncountable set. We will use the Schroeder - Bernstein Theorem:
$f: \mathbb{R} \setminus \mathbb{N} \to \mathbb{R}$, $f(x) = x$
$g:\mathbb{R} \to \mathbb{R} \setminus \mathbb{N}$, $g(x) = \frac{1}{e^{x} + 2}$
As we have defined two injections, by the Schroeder-Bernstein theorem there exists a bijection between $\mathbb{R} \setminus \mathbb{N}$ and $\mathbb{R}$. Thus, $\mid \mathbb{R} \setminus \mathbb{N} \mid = \mid \mathbb{R} \mid$, so $\mathbb{R} \setminus \mathbb{N}$ is uncountable. Thus, $R \setminus A$ is uncountable. $\square$

Answer 1

If the complement was countable then the set of reals would be the union of two countable sets, hence countable, a contradiction.