 Solution

The rational cut is both open and closed in $\mathbb{Q}$. It is relatively open as $\mathbb{Q}\cap (-\infty ,\sqrt{2})$. Since $\sqrt{2}$ is irrational, its relative complement is $\mathbb{Q}\cap (\sqrt{2},\infty )$, also relatively open. Thus this cut is a clopen set in the subspace topology of the rationals.