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).
It remains only to show that $H^0(L^{\otimes n} |_Y)\neq 0$ implies $H^0(L^{\otimes kn})|_Y\neq 0$, for some $k>0$ depending on the codimension of $Y$.
If $Y$ does not lie in the zero divisor of $L^{\otimes n}$, we are done, because $H^0(L^{\otimes n})|_Y\neq 0$. Otherwise we use induction in codimension of $Y$, reducing the statement to the situation when $Y$ is a component of the zero divisor of a section of $L^{\otimes n}$. Then we may replace $Y$ with the zero divisor of a section of $L^{\otimes n}$, and obtain an exact sequence $$ 0\rightarrow H^0(L^{\otimes n}) \rightarrow H^0(L^{\otimes 2n}) \stackrel r \rightarrow H^0(Y, L^{\otimes n} |_Y)\rightarrow 0 $$ Since the rightmost arrow $r$ is just the restriction map, this implies that $H^0(L^{\otimes 2n})|_Y$ is non-zero.