Solution: To prove that $\mathbb{R} \setminus \mathbb{Q} $ is separable, we need to find $A \subseteq \mathbb{R} \setminus \mathbb{Q} $ such that $A$ is dense in $\mathbb{R} \setminus \mathbb{Q} $, that is, $\bar{A} = \mathbb{R} \setminus \mathbb{Q} $. Consider the set \[ A = \left\{ \frac{\sqrt{2} p}{q} : p,q\in \mathbb{Z},\ q\neq 0 \right\}. \] It is clear that $A$ is countable. We claim that $A$ is dense in $\mathbb{R} \setminus \mathbb{Q} $. For this, we will show that given any irrational number $x$, we can find a sequence of numbers from $A$ converging to $A$. Let $\tilde{x} = \frac{x}{\sqrt{2} }$. We know that given any real number $a$, there exists a sequence of rational numbers converging to $a$. Therefore, we can find a sequence of rational numbers, say $\left\{ \frac{p_n}{q_n} \right\} $, converging to $\tilde{x} $. Thus, \[ \frac{p_n}{q_n} \rightarrow \frac{x}{\sqrt{2} } \implies \sqrt{2} \cdot \frac{p_n}{q_n} \to x. \] Thus, we found a sequence, namely $\left\{ \frac{p_n\sqrt{2} }{q_n} \right\} $, converging to $x$, and therefore, $A$ is dense in $\mathbb{R} -\mathbb{Q} $.