The proof is actually pretty easy. Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).
However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.