Curves with negative self-intersection

Question

Let $S$ be a non-singular projective surface over $\mathbb{C}$. Show that $S$ contains at most countably many irreducible curves $C$ with $C^2<0$.

Answer 1

The key point is that if $C$ and $C'$ are distinct irreducible curves, then $C \cdot C' \geq 0$. Therefore if $C$ is an irreducible curve with $C^2<0$, it is the unique such curve in its numerical equivalence class.

Now the group $N^1(S)$ of numerical equivalence classes of curves on $S$ is a finitely-generated (hence countable) abelian group, by the "theorem of the base" of Severi and Néron.

Answer 2

As a complement to Asal's perfect answer let me remark that it is indeed possible for a smooth projective surface to carry infinitely many curves with negative self intersection.
Indeed, by blowing-up nine general points in $\mathbb P^2$ one obtains a surface $S$ containing infinitely many $(-1)$-curves i.e. curves $C$ isomorphic to $\mathbb P^1$ and such that $C\cdot C=-1$.
Apparently this was claimed by Franchetta who gave an incomplete argument and then proved by Nagata under an indication by Kodaira: see the original (and difficult to read) article here.
A more user-friendly proof can be found on page 131, Proposition 22 of Friedman's book.