I hope nobody would mind if I try to do the exercise.

Clearly f is continuous on G. Suppose f is continuous on x and f(x)=1/k. Take epsilon=1/k. Let U be any neighborhood of x. U\cap G_1\cap .. \cap G_{k-1} contains an irrational number y. Hence |f(x)-f(y)|=2/k > epsilon. (if f(x)=-1/k, take y to be a rational number)

I hope nobody would mind if I try to do the exercise.

Clearly $f$ is continuous on $G$. Suppose $f$ is continuous on $x$ and $f\left(x\right)=1/k$. Take $\epsilon=1/k$. Let $U$ be any neighborhood of $x$. $U\cap G_1\cap .. \cap G_{k-1}$ contains an irrational number $y$. Hence $\left|f\left(x\right)-f\left(y\right)\right|=2/k>\epsilon$. (If $f\left(x\right)=-1/k$, take $y$ to be a rational number.)