The proof is actually very easy.
From your assumption, 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, positively-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 vartiety
is projective, which can be seen from vanishing
of cohomology of powers of $L$ (a finite
map is acyclic on coherent sheaves, hence
cohomology of $L^{\otimes n}$ on $X$ are the
same as cohomology of ${\cal O}(1)$
on its image).