Let $S$ be a complex projective smooth surface. If $D$ is a divisor on $S$, let's write $h^i(D)$ for $dim H^i(S,\mathcal{O}_S(D))$, where $\mathcal{O}_S(D)$ is the invertible sheaf associated to $D$. Let's denote with $K$ a canonical divisor, so $K$ is such that $\mathcal{O}_S(K)=\Omega^2_S$.
I want to prove Castelnuovo's rationality criterion, so let $S$ be a surface such that $q(S)=P_2(S)=0$. I've already proved that with this assumption, there is on $S$ a smooth rational curve $C$ such that $C^2\geq 0$. From this using Riemann-Roch and the genus formula, i get $\chi(\mathcal{O}_S(C ))\geq 2$.
Now i want to show that $h^0(C )\geq 2$. To do this it suffices to show that $h^2( C)=0$.
My question is: is it correct to say that $h^2( C)=h^0(K-C)\leq h^0(K)$?
Yes everything you write is correct.
In particular your very last inequality follows from the implication for divisors on $S$ (or on any smooth variety for that matter):
$$ D\leq E\implies H^0(S, \mathcal O(D))\subset H^0(S, \mathcal O(E)) $$ This implication is evident by interpreting $H^0(S, \mathcal O(D))$ as the vector space of rational functions $f$ on $S$ with $\text {div}(f)+D\geq 0$ .
In your case you are applying this to $K-C\leq K$ .