Separatedness is local on the base

Put $Q_i=(X{\times }_{S}X){\times }_S{S}_i$, the inverse image of $S_i$ under the structure morphism $X{\times }_{S}X\to S$. The $Q_i$ are open subschemes of $X{\times }_{S}X$ and they cover $X{\times }_{S}X$, because the $S_i$ cover $S$.

From the Cartesian square of [F2] and the identity ${\mathrm{pr}}_j{\mathrm{\Delta }}_{X\mathrm{/}S}={\mathrm{id}}_X$ of the diagonal, the inverse image ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q_i)$ is exactly $X_i$ and the restriction of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ to $Q_i$ is the diagonal ${\mathrm{\Delta }}_{X_i\mathrm{/}{S}_i}$.

By [F2] applied to $S_i\to S$ there is a canonical isomorphism $X_i{\times }_{S_i}{X}_i\cong Q_i$ under which ${\mathrm{\Delta }}_{X_i\mathrm{/}{S}_i}$ is the base change of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ along the open immersion $Q_i\to X{\times }_{S}X$.

Conversely assume each $f_i$ is separated, so that each diagonal ${\mathrm{\Delta }}_{X_i\mathrm{/}{S}_i}$ is a closed immersion by [F1]. By step 1.2 the restriction of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ to the open subscheme $Q_i$ is exactly ${\mathrm{\Delta }}_{X_i\mathrm{/}{S}_i}$ and ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q_i)=X_i$, so every restriction of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ to a member of the open cover $\{Q_i\}$ of $X{\times }_{S}X$ of step 1.1 is a closed immersion.

If $f$ is separated, then ${\mathrm{\Delta }}_{X\mathrm{/}S}$ is a closed immersion by [F1], and its base change ${\mathrm{\Delta }}_{X_i\mathrm{/}{S}_i}$ along the open immersion $Q_i\to X{\times }_{S}X$ is a closed immersion by [F4]; hence each $f_i$ is separated by [F1].

By [F3] applied to the cover $\{Q_i\}$ of $X{\times }_{S}X$, the diagonal ${\mathrm{\Delta }}_{X\mathrm{/}S}$ is a closed immersion, so $f$ is separated by [F1].

Steps 3.1 and 3.2 give both implications, including the cases where some $S_i$ or $X_i$ is empty and the case of a one-element cover.