I am going to use the sequential definition of continuity and show the definition fails at rational and irrational points. It is not continuous anywhere. Let be rational. Let , which is a sequence of irrational numbers which converges to . Since each is irrational, for all positive integers , so it trivially converges to , and since , we have that yet , so the function is discontinuous on the rationals. Now let be irrational. By the denseness of , we can construct a sequence of rational numbers that converges to . Since each is rational, we have that for all positive integers . So trivially converges to , and since , we have that yet , so the function is discontinuous on the irrationals.
If you need to find the values using the epsilon-delta definition of continuity, the basic idea behind the proof is that any interval contains infinitely many rationals and irrationals, so if the function were continuous then must get arbitrarily close to 0 and 2 simultaneously, which is clearly impossible.