Let $X$ be a surface (therefore $2$-dimensional, proper $k$-scheme) and $D$ a divisor with positive self intersection number $(D \cdot D) >0$. Futhermore it is nef therefore for each irreducible curve $C$ we have $(D \cdot C) \ge 0$.
Is then $D$ semi ample? By semi ampleness I mean here that there exist a semi ample invertible sheaf $\mathcal{L}$ (so for exery $x \in X$ there exist $n \in \mathbb{N}$ and a global section $s \in \Gamma(X, \mathcal{L}^{\otimes n}$) with property $s(x) \neq 0$)
such that $\mathcal{L} = \mathcal{O}_X(D)$ holds.
Remark: Since we demand only $(D \cdot C) \ge 0$ and not $(D \cdot C) > 0$ we can't apply Nakai-Moishezon.
Best Answer
No, this is not true. Note that for $D$ nef, the condition that $D^2>0$ is equivalent to $D$ being big.
There is a famous example due to Zariski of a surface $S$ and a divisor $D$ on $S$ such that $D$ is nef and big, but $D$ is not semi-ample. Roughly, we construct $S$ by blowing up $\mathbf P^2$ in a certain set of 12 points on an elliptic curve $C$. The fact that there are non-torsion line bundles of degree 0 on $C$ allows us to choose the points in such a way that the line bundle $D=4H-\sum_i E_i$ on $S$ is nef and big, but for every $m>0$ the proper transform of $C$ is in the base locus of $|mD|$.
For details, see Section 2.3.A of Positivity in Algebraic Geometry by Lazarsfeld.