Continuous function on $\mathbb{Q}$

Because $\sqrt{2}$ is not in $\mathbb{Q}$, you don't have to check continuity at $\sqrt{2}$: it's completely irrelevant!

Hint: You are right, you need only check continuity around (i.e. nearby) $\sqrt 2$, not at $\sqrt 2$.

$f$ is continuous if it is continuous in every point in $\mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.

Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.