Answer
There are no points where $f(x)$ is continuous.
Work Step by Step
Let's assume that $f(x)$ is continuous at $x = c$
Let $\epsilon = \frac{1}{2}$
There is a positive number $\delta$ such that if $\vert x-c \vert \lt \delta,$ then $\vert f(x) - f(c) \vert \lt \epsilon$
Choose a rational number $a$ in the interval $(c-\delta, c + \delta)$
Then $\vert a-c \vert \lt \delta,$ so $\vert f(a) - f(c) \vert \lt \epsilon$
Choose an irrational number $b$ in the interval $(c-\delta, c + \delta)$
Then $\vert b-c \vert \lt \delta,$ so $\vert f(b) - f(c) \vert \lt \epsilon$
Note that $f(a) = 0$ and $f(b) = 1$
Then:
$\vert f(a)-f(b)\vert = \vert f(a)-f(c)+f(c)-f(b)\vert \leq \vert f(a)-f(c)\vert + \vert f(c)-f(b) \vert \lt \epsilon+\epsilon = 1$
Clearly, this is a contradiction since $\vert f(a)-f(b)\vert = 1$
Our initial assumption must be false. Therefore, there are no points where $f(x)$ is continuous.
