Dirichlet function

Counterexample

Let $f \colon \mathbb{R} \to \mathbb{R}$ be given by $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$. Then $f$ is not continuous at any point. Indeed, each real $a$ is the limit of a sequence of rationals and of a sequence of irrationals. Their images tend to $1$ and to $0$, so one of them does not tend to $f(a)$.

Remarks

The function $f$ is the indicator function of $\mathbb{Q}$. Every rational number is a period of $f$. The function is not Riemann integrable on any interval with more than one point.

Explanations

L1 human

Rational and irrational numbers lie mixed together in every small interval. So the function jumps between $0$ and $1$ near every point, and it has no limit anywhere.