I think the following is a countable dense subset of irrational numbers. I request you to verify this.
Since $\Bbb R -\Bbb Q$ is dense in $\Bbb R$, for every rational number $r$ there is a sequence of irrational numbers, which I will denote by $(\alpha_k^r),k\in \Bbb N$ such that $lim(\alpha_k^r) = r$. For a fixed $r\in\Bbb Q$, the set of points
$ \{ (\alpha_k^r)|k\in\Bbb N \} $ must be infinite, otherwise some irrational number would be equal to $r$. Also $r\in\overline {\{ (\alpha_k^r)|k\in\Bbb N \}} $. Now consider the set, $$\Gamma = \bigcup_{r\in\Bbb Q} \{ (\alpha_k^r)|k\in\Bbb N \}\subset \Bbb R-\Bbb Q$$ which is a countable union of countable sets and hence countable. And because of the previous remarks, $\overline \Gamma = \Bbb Q$ . But $\overline{ \Bbb Q} = \Bbb R$ and $\overline{\overline \Gamma} = \overline\Gamma .$ So, $\overline\Gamma = \Bbb R. $
Best Answer
Your proof is correct, though it could be directer: $f(x) = x + \sqrt{2}$ is continuous from $\mathbb{R}$ to itself.
So $\mathbb{Q}$ being dense in $\mathbb{R}$ implies that $f[\mathbb{Q}]$ is dense in $\mathbb{R}$ as well. And this is a set of irrationals, as $q + \sqrt{2} \in \mathbb{Q}$ implies $\sqrt{2} \in \mathbb{Q}$, quod non.
We could also have used $f(x) = x\sqrt{2}$ for $\mathbb{R}\setminus \{0\}$, of course. Etc. etc.