Let $X$ be an algebraic curve, $D$ a divisor and $\mathscr{O}(D)$ the line bundle associated to $D$ in the canonical way. The following implication should follow immediately from the Riemann Roch formula
$$deg(D)<0 \implies h^0(X, \mathscr{O}(D)) =0.$$
Could you help me to see why this is the case?
This has nothing to do with Riemann-Roch:
The vector space of global sections $H^0(X, \mathscr{O}(D)) $ is equal to the set of rational functions $f\in \operatorname{Rat}(X)$ for which the divisor $E=\operatorname{div}(f)+D$ is effective.
Effectivity of $E$ implies $ \deg(E)=\deg(\operatorname{div}(f))+\deg(D) \geq 0$.
Since for a non-zero $0\neq f$ we have $\deg(\operatorname{div}(f))=0$, we see that if a non-zero $f\in H^0(X, \mathscr{O}(D)) $ existed we would have $ \deg D \geq 0$.
Contrapositively the assumption $\deg(D)\lt 0$ forces $H^0(X, \mathscr{O}(D)) =0$, just as you required.