I understand this proof that $\mathbb{Q}$ is countable:
The rational numbers are arranged thus: $$\displaystyle \frac 0 1, \frac 1 1, \frac {-1} 1, \frac 1 2, \frac {-1} 2, \frac 2 1, \frac {-2} 1, \frac 1 3, \frac 2 3, \frac {-1} 3, \frac {-2} 3, \frac 3 1, \frac 3 2, \frac {-3} 1, \frac {-3} 2, \frac 1 4, \frac 3 4, \frac {-1} 4, \frac {-3} 4, \frac 4 1, \frac 4 3, \frac {-4} 1, \frac {-4} 3 \ldots$$ It is clear that every rational number will appear somewhere in this list. Thus it is possible to set up a bijection between each rational number and its position in the list, which is an element of $\mathbb N$. $\square$
However, given any rational number, say $3$, there is no "next" rational number, as $\mathbb Q$ is dense in $\mathbb R$. But, given any natural number, we can determine the next natural number (i.e. its successor, given by $S(n)=n+1$).
This seems to contradict the fact that there's a bijection between $\mathbb Q$ and $\mathbb N$. Could someone clear this up?
(Forgive me for any misconceptions/mistakes- I'm an amateur in set theory)