Yes, the function is continuous at 0. Given any > 0, [tex]|f(x)- f(0)| is |x- 0|= |x| if x is rational, and |-x- 0|= |x| if x is irrational. Take to be any number less than .
If a is any rational number other than 0, f(x)= a so |f(x)- f(a)= |x-a| if x is rational and |-x-a|= |x+a| if x is irrational. No matter how small is, there will exist both rational and irrational numbers between and so with less than half the difference between them, there will exist x such that |f(x)- f(a)|> [tex]\epsilon[/b]. The same is true if a is irrational.