Let $s \in H^0(X, \mathcal{L}^{\otimes d})$ be a global section. Let $U \subset X$ be the open subspace over which $s$ is a generator of $\mathcal{L}^{\otimes d}$. In particular we have $\mathcal{L}^{\otimes d}|_U \cong \mathcal{O}_U$. We claim that $U$ is affine. \medskip\noindent Proof of the claim. We will show that $H^1(U, \mathcal{F}) = 0$ for every quasi-coherent $\mathcal{O}_U$-module $\mathcal{F}$. This will prove the claim by Proposition \ref{proposition-vanishing-affine}. Denote $j : U \to X$ the inclusion morphism. Since \'etale locally the morphism $j$ is affine (by Morphisms, Lemma \ref{morphisms-lemma-affine-s-open}) we see that $j$ is affine (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-affine-local}). Hence we have $$ H^1(U, \mathcal{F}) = H^1(X, j_*\mathcal{F}) $$ by Lemma \ref{lemma-affine-vanishing-higher-direct-images} (and Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-apply-Leray}). Write $j_*\mathcal{F} = \colim \mathcal{F}_i$ as a filtered colimit of coherent $\mathcal{O}_X$-modules, see Lemma \ref{lemma-directed-colimit-coherent}. Then $$ H^1(X, j_*\mathcal{F}) = \colim H^1(X, \mathcal{F}_i) $$ by Lemma \ref{lemma-colimits}. Thus it suffices to show that $H^1(X, \mathcal{F}_i)$ maps to zero in $H^1(U, j^*\mathcal{F}_i)$. By assumption there exists an $n \geq 1$ such that $$ H^1(X, \mathcal{F}_i \otimes_{\mathcal{O}_X} (\mathcal{O}_X \oplus \mathcal{L} \oplus \ldots \oplus \mathcal{L}^{\otimes d - 1}) \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n}) = 0 $$ Hence there exists an $a \geq 0$ such that $H^1(X, \mathcal{F}_i \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes ad}) = 0$. On the other hand, the map $$ s^a : \mathcal{F}_i \longrightarrow \mathcal{F}_i \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes ad} $$ is an isomorphism after restriction to $U$. Contemplating the commutative diagram $$ \xymatrix{ H^1(X, \mathcal{F}_i) \ar[r] \ar[d]_{s^a} & H^1(U, j^*\mathcal{F}_i) \ar[d]^{\cong} \\ H^1(X, \mathcal{F}_i \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes ad}) \ar[r] & H^1(U, j^*(\mathcal{F}_i \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes ad})) } $$ we conclude that the map  $H^1(X, \mathcal{F}_i) \to H^1(U, j^*\mathcal{F}_i)$ is zero and the claim holds. \medskip\noindent Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma \ref{decent-spaces-lemma-decent-space-closed-point} we can represent $x$ by a closed immersion $i : \Spec(k) \to X$ (this also uses that a quasi-separated algebraic space is decent, see Decent Spaces, Section \ref{decent-spaces-section-reasonable-decent}). Thus $\mathcal{O}_X \to i_*\mathcal{O}_{\Spec(k)}$ is surjective. Let $\mathcal{I} \subset \mathcal{O}_X$ be the kernel and choose $d \geq 1$ such that $H^1(X, \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes d}) = 0$. Then $$ H^0(X, \mathcal{L}^{\otimes d}) \to H^0(X, i_*\mathcal{O}_{\Spec(k)} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes d}) = H^0(\Spec(k), i^*\mathcal{L}^{\otimes d}) \cong k $$ is surjective by the long exact cohomology sequence. Hence there exists an $s \in H^0(X, \mathcal{L}^{\otimes d})$ such that $x \in U$ where $U$ is the open subspace corresponding to $s$ as above. Thus $x$ is in the schematic locus (see Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}) of $X$ by our claim. \medskip\noindent To conclude that $X$ is a scheme, it suffices to show that any open subset of $|X|$ which contains all the closed points is equal to $|X|$. This follows from the fact that $|X|$ is a Noetherian topological space, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-Noetherian-sober}. Finally, if $X$ is a scheme, then we can apply Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-compact-h1-zero-invertible} to conclude that $\mathcal{L}$ is ample.