Suppose $f(x) = \begin{cases} 0 \ \ \text{if} \ x \ \text{is irrational} \newline \frac{1}{n} \ \ \text{if} \ x = m/n \ \text{in lowest form} \end{cases}$
To show that $f$ is continuous at all irrational numbers in $(0,1)$ let $a$ be an irrational in the interval. Let $\epsilon$ be given. Choose $q$ such that $1/q < \epsilon$. Why are thee are at most $1+2+ \cdots + (q-1)$ rationals in $(0,1)$ of the form $m/n$ where $n
