Accumulation points of $\mathbb{Q}$

Proof. Let $x\in\mathbb{R}$ and $\varepsilon>0$. Since $\mathbb{Q}$ is dense, there is $q\in\mathbb{Q}\cap(x-\varepsilon,x+\varepsilon)$. If $q\neq x$, done; if $q=x$, take another rational in the smaller ball $(x-\varepsilon,x+\varepsilon)\setminus\{x\}$.

Conclusion. Every real number is an accumulation point of $\mathbb{Q}$; $\overline{\mathbb{Q}}=\mathbb{R}$.