Consider $ f : \mathbb{R} \to \mathbb{R} $ with $ f(x) = 0 $ at irrational $ x $ and $ f(\frac{p}{q}) = \frac{1}{q} $ at rationals. This is Thomae’s function.

Btw when we say “a rational $ \frac{p}{q} $” it will be implicit that $ p, q $ are integers, $ q > 0 $ and the fraction is in reduced form.

$ f $ is discontinuous at rationals.

If it were continuous at some $ \frac{p}{q} $, there will be a neighborhood $ (\frac{p}{q}-\delta , \frac{p}{q} + \delta) $ on which the function is positive. But every nbhd has irrationals and at these function takes value $ 0 $, absurd.

$ f $ is continuous at every irrational $ x $.

Fix irrational $ x $. Let $ \epsilon > 0 $. We need a $ \delta > 0 $ such that function values on $ (x-\delta, x+\delta) $ are of magnitude $ < \epsilon $.
Pick integer $ Q > 0 $ with $ \frac{1}{Q} < \epsilon $. Now it suffices to pick a $ \delta > 0 $ such that all rationals in $ (x-\delta, x+\delta) $ have denominators $ > Q $ : so that at rational points, 1/denom < 1/$ Q $ < $ \epsilon $.
Imagine grids $ \frac{1}{q} \mathbb{Z} $ for integers $ q > 0 $, $ x $ cant be on any of these. Also each of $ \frac{1}{1} \mathbb{Z}, \frac{1}{2} \mathbb{Z}, \ldots, \frac{1}{Q} \mathbb{Z} $ has finite intersection with $ (x-1, x+1) $. So we can pick a $ 0 < \delta < 1 $ such that $ (x-\delta, x+\delta) $ avoids these intersection points, and now rationals in $ (x-\delta, x+\delta) $ in reduced form cant have denominator $ \leq Q $, as needed.