Show that the closure of $\Bbb{Q}$ is equal to $\Bbb{R}$

Question

Proof (Show:$\Bbb{\overline{Q}}=\Bbb{R}$): Consider that $\Bbb{\overline{Q}}:=\{r\in\Bbb{R}:r\in\Bbb{Q}\quad \lor\quad \forall\delta>0\exists q\in\Bbb{Q}(\vert r-q\vert<\delta)\}$. Certainly, $\Bbb{\overline{Q}}\subseteq \Bbb{R}$. So we are left to show $\Bbb{R}\subseteq\Bbb{\overline{Q}}$. Let $x\in\Bbb{R}$ and $y\in\Bbb{Q}$. By a known result, since $x,y\in\Bbb{R}$, then $\exists q\in\Bbb{Q}$ such that $x<q<y$. Since $x$ and $y$ are arbitrary real numbers, then $\vert x-y\vert=\delta>0$ is also arbitrary. Note that \begin{align} q<y &\Leftrightarrow -q<-y\\ &\Leftrightarrow x-q>x-y\\ &\Leftrightarrow -(x-q)<-(x-y)\\ &\Leftrightarrow \vert x-q\vert<\vert x-y\vert =\delta \end{align} Thus, $\forall\delta>0,\exists q\in\Bbb{Q}$, $\vert x-q\vert<\delta$ and this implies $x\in\Bbb{Q}$ and $\Bbb{R}\subseteq\Bbb{\overline{Q}}$. Therefore, $\Bbb{R}=\Bbb{\overline{Q}}$

Answer 1

You wrote that $x\in\mathbb R$ and that $y\in\mathbb Q$, but after you wrote that $x$ and $y$ are arbitrary real numbers. This is a contradiction.

If $x\in\mathbb R$ and $\delta>0$, take a rational number $q\in(x-\delta,x+\delta)$. Then $|x-q|<\delta$ and you're done.

Answer 2

In the set you've defined as $\mathbb{Q}$ there are irrational numbers, since they are also dense in $\mathbb{R}$.

Another thing that feels weird is trying to prove $\mathbb{Q}\subseteq \overline{\mathbb{R}}$ when $\overline{\mathbb{R}}=\mathbb{R}$, which already contains the rational numbers. I think you meant $\mathbb{R}\subseteq\overline{\mathbb{Q}}$.

In general, I think the wording could be better, but the idea is clear, you have to show that for any real number $t$ and any $\delta>0$, there is a rational $q$ such that $t\in B(q,\delta)$, but that is inmediate by the known result you mention.