 Solution

The rational sequence $q_n=10^{-n}\lfloor 10^n\sqrt{2}\rfloor$ satisfies $0\le \sqrt{2}-q_n<10^{-n}$. It is therefore a Cauchy sequence in the induced metric on $\mathbb{Q}$. Any limit in $\mathbb{Q}$ would also be its real limit, which is $\sqrt{2}\notin \mathbb{Q}$. ThusCompleteness concerns limits lying in the space, so being a dense subset of $\mathbb{R}$ does not suffice.