(From Vakil's notes, Exercise 18.4.K)
If $C$ is an integral projective curve over a field $k$, and $\mathscr{L}$ is an ample line bundle on $C$, why is the degree of $\mathscr{L}>0$?
If $C$ is also a regular curve, then I think I can do this, as it can be quickly checked that ${\rm deg}(\mathscr{L}^{\otimes n}) = n {\rm deg}(\mathscr{L})$ and, if $\mathscr{L}$ is very ample, there exists a global section (and that section must vanish somewhere or else our very ample bundle is trivial).
However, I don't see how to extend this argument to the case where the curve might not be regular. The degree of a coherent sheaf is defined in the notes as
\begin{align*}
{\rm deg}(\mathscr{F}):=\chi(C,\mathscr{F}) – ({\rm rank} \mathscr{F})\chi(C,\mathscr{O}_C)
\end{align*}
As one possible useful fact, Exercise 18.4.S seems to say that we can express $\mathscr{L}$ as $\mathscr{O}_C(\sum{n_ip_i})$ where $p_i$ are regular points that are not associated points. I don't know why this is true, but it would at least allow us to view $\mathscr{L}$ more concretely (in particular, repeating the argument for Riemann Roch for regular projective curves tells us the degree of $\mathscr{L}$). However, even assuming this fact, I would need to know be able to compute the degree of $\mathscr{L}$ using the zeros and poles of a global section to use the argument above.
Best Answer
The result and your argument (which is the first I would think of) are still valid for integral projective curves, but the theory is more delicate. For example, you can find the result as 7.3.2 Proposition 3.25(b) of Liu's wonderful book, in which he develops Riemann-Roch results with minimal assumptions on what constitutes a "curve" (e.g. possibly singular, nonreduced).