Must a proper curve minus a point be affine?

Theorem
If a nonzero finite number of points $p_1,\dots, p_r $ are deleted from $C$ the resulting curve will be affine.

Indeed consider the divisor $D=p_1+\dots+ p_r $ on $C$.
Since it has positive degree some positive multiple $nD$ of it will be very ample.
Thus we get an embedding of $j:C\to \mathbb P^N$ (for some huge $N$) and a hyperplane section divisor $\Delta =H\cap j(C)$ on $j(C)$ such that $j^*\Delta=nD$.
But then $C\setminus \{p_1,\dots, p_r\}$ is isomorphic to $j(C)\cap (\mathbb P^N\setminus H)\cong j(C)\cap \mathbb A^N$ (the complement of a hyperplane in projective space is affine space) and since this last variety $j(C)\cap \mathbb A^N$ is clearly affine, so is $C\setminus \{p_1,\dots, p_r\}$.

Edit
The theorem is valid even if $C$ is singular.
To see that, consider the finite normalization morphism $n:\tilde C\to C$ and delete the inverse image of $\{p_1,\dots, p_r\}$, obtaining the smooth curve $C'=\tilde C\setminus n^{-1}(\{p_1,\dots, p_r\})$ which is affine by the result already proved for smooth curves.
Now consider the restricted finite morphism $n':C'\to C\setminus \{p_1,\dots, p_r\}$.
Since $C'$ is affine and the finite morphism $n'$ is surjective the curve $C\setminus \{p_1,\dots, p_r\}$ will also be affine by Chevalley's Theorem (EGA $_{II}$, Théorème (6.7.1), page 136), and we are done.