It is enough to prove that every prime cycle is integrable. Applying the Chow Lemma to the support of the cycle and using that the inverse image of a quasi-algebraic metric is quasi-algebraic, we are reduced to the case when $X$ is projective.

We proceed by induction on $d$. For $d=-1$, the statement is clear, and so we consider the case when $d\geq 0$. Let $Y$ be a $d$-dimensional cycle of $X$ and $s_{i}$, $i=0,\dots,d$, rational sections of $L_{i}$ that intersect $Y$ properly. Let $({\mathcal{X}},{\mathcal{L}}_{d})$ be a proper model over $\mathbb{K}^{\circ}_{S}$ of $(X,L_{d}^{\otimes e_{d}})$. Then $s_{d}^{\otimes e_{d}}$ is a non-zero rational section of ${\mathcal{L}}_{d}$ and so it defines a finite number of vertical components. Hence, for all places $v\notin S$ which are not below any of these vertical components,

thanks to the equation (2.43). The statement follows then from the inductive hypothesis. ∎