 Solution

At a nonzero rational $p/q$ in lowest terms, irrational points approach $p/q$ with function value zero, whereas $f(p/q)=1/q>0$, so $f$ is discontinuous. At zero, $|p/q|\ge 1/q$ for nonzero $p$, hence $0\le f(x)\le |x|$ and continuity follows.

At an irrational $x$, given $\epsilon >0$, only finitely many rationals in a bounded neighbourhood have reduced denominator $q\le 1/\epsilon$. Choose a neighbourhood avoiding those points; throughout it, $f<\epsilon$. Consequently the continuity set is $(\mathbb{R}\setminus \mathbb{Q})\cup \{0\}$. This is a variant of the Thomae function.