No it's not correct. The function is not continuous at any point. (you were right on this).

And for $\displaystyle \lim_{n\to \infty} f(\frac{1}{n})$ doesn't exist. It's equivalent to write $\displaystyle \lim_{x \to 0^+} f(x)$. When x tends to 0, f(x) doesn't tend to anything, it get the values 0 and 1 again and again and it doesn't tend to 1 as you said.

And about I think the implication doesn't work. For example if you have the function $\displaystyle f(x)=x$, then you could find both a rational and irrational sequence converging to, say 3. But the functions is continuous at any point. So be careful.