The classical example of a functions with only one point of continuity is $ f(x) = \begin{cases} x & \text{if } x \in \mathbf{Q}, \\ 0 & \text{otherwise}. \end{cases} $
I only want to prove the continuity at $0$, using the definition in terms of open sets. Let $(a,b)$ be an open set containing $0$. I want to prove that the preimage of this interval is open in $\mathbf R$. But since every preimage contains the $0$ element, the preimage consists of all irrational numbers and the rational numbers of the interval $(a,b)$, but this set is not open. What am I doing wrong?