I would like to know it the following proof is correct:
Lemma: Let $C$ be a smooth, irreducible curve of genus $g$ over a complex projective K3 surface $X$. Let $L:=\mathcal{O}_X(C)$. Then $C^2=2g-2$ and $h^0(X,L)=g+1$.
Proof: For the first equality it suffices to use the genus formula. Then, by Serre duality, since $X$ is K3, we have $h^1(X,L)=h^1(X,-L)$ and $h^2(X,L)=h^0(X,-L)=0$ (because $C$ is irreducible). By Riemann Roch and by the equality $C^2=2g-2$, we get $h^0(X,L)-h^1(X,L)+h^2(X,L)=g+1$. Hence it suffices to show the equality $h^1(X,-L)=0$. The usual exact sequence
$0 \to \mathcal{O}_X(-C) \to \mathcal{O}_X \to \mathcal{O}_C\to 0 $
induces the exact sequence in cohomology
$0 \to H^0(X,-C) \to H^0(X,\mathcal{O}_X) \to H^0(X,\mathcal{O}_C) \to H^1(X,-C) \to H^1(X,\mathcal{O}_X)=0 $
We know $h^0(X,-C)=0$, $h^0(X,\mathcal{O}_X)=1$ (since $X$ is compact), $h^0(X,\mathcal{O}_C)=1$ (since $C$ is a compact Riemann Surface) and the map
$H^0(X,\mathcal{O}_X) \to H^0(X,\mathcal{O_C})$ is just the identity. Thus i can conclude that $h^1(X,-C)=0$
Note that with $\mathcal{O}_C$ i mean $i_{*}\mathcal{O}_C$, where $i \colon C \hookrightarrow X$.
Do you think it works?