This fact without a proof is mentioned on the page 147 of Bogomolov's book "Algebraic curves and onedimensional fields", but the statement is not evident to me. Can anyone explain to me why is it true? Thank you very much.
If $X$ is a smooth projective curve and $P\in X$ is any point, then $X\setminus P$ is affine. Indeed, by RiemannRoch for $m$ large enough the linear system $mP$ embeds $X$ in some projective space $\mathbb P^n$, so there is a hyperplane $H_0$ of $\mathbb P^n$ that intersects $X$ only at $P$. So, if $P\ne Q$ then $X_1=X\setminus\{P\}$ and $X_2=X\setminus\{Q\}$ are affine curves such that $X=X_1\cup X_2$.
One way to see this would be as follows: if $X\subset\mathbb{P}^n$ is a curve one can find hyperplanes $H_1,H_2$ such that $X\cap H_1\cap H_2=\varnothing$. Then $X$ will be the union of two affine curves $X\setminus H_1$ and $X\setminus H_2$.

2$\begingroup$ Very nice ! Over a finite field, one can use hypersurfaces instead of hyperplanes. $\endgroup$ – Qing Liu Nov 17 '12 at 22:09

1$\begingroup$ Qing Liu  yes, indeed. By the way, slightly generalizing the argument one sees that a projective variety $X$ can be covered by $\dim X+1$ affine sets. $\endgroup$ – algori Nov 17 '12 at 23:52