The empty morphism is finite, proper and projective

The empty scheme is $\mathrm{Spec}0$ by [F1], so it is a scheme and the empty morphism $\mathrm{\varnothing }\to S$ is a morphism of schemes. For every affine open $U=\mathrm{Spec}A\subseteq S$ the inverse image is empty, that is $\mathrm{Spec}0$, and the zero ring is module-finite over $A$ by [F3]. Hence $\mathrm{\varnothing }\to S$ is finite by [F2].

The morphism is of finite type: it is locally of finite type because the empty inverse image $\mathrm{Spec}0$ of every affine open is affine with coordinate ring $0$, which is a finitely generated $A$-algebra by [F3], and it is quasi-compact because an empty inverse image is covered by the empty finite subcover, so the condition of [F4] holds for every quasi-compact open of $S$.

The morphism is separated. Its diagonal is a morphism $\mathrm{\Delta }:\mathrm{\varnothing }\to \mathrm{\varnothing }{\times }_S\mathrm{\varnothing }$; the fibre product of two empty schemes is empty, since both projections would have to map into the empty scheme. Thus $\mathrm{\Delta }$ is the empty morphism $\mathrm{\varnothing }\to \mathrm{\varnothing }$, whose underlying map is a homeomorphism onto the closed subset $\mathrm{\varnothing }$ of $\mathrm{\varnothing }$ and whose structure map ${\mathcal{O}}_{\mathrm{\varnothing }}\to {\mathrm{\Delta }}_{\ast }{\mathcal{O}}_{\mathrm{\varnothing }}$ has zero target, hence is surjective. By [F5], $\mathrm{\Delta }$ is a closed immersion, so [F6] makes $\mathrm{\varnothing }\to S$ separated.

The morphism is universally closed. For any $S$-scheme $T$ the base change $\mathrm{\varnothing }{\times }_{S}T$ is empty, because the projection to $\mathrm{\varnothing }$ must map into the empty scheme; the base-changed projection therefore has empty domain, and the image of its only closed subset $\mathrm{\varnothing }$ is $\mathrm{\varnothing }$, which is closed in $\mathrm{\mid }T\mathrm{\mid }$. Hence the condition of [F7] holds for every $T$, and $\mathrm{\varnothing }\to S$ is universally closed.

Steps 1.2, 1.3 and 1.4 give finite type, separatedness and universal closedness, so $\mathrm{\varnothing }\to S$ is proper by [F8].

Since [F10] gives ${\mathbb{P}}_S^0\cong S$, the unique morphism $i:\mathrm{\varnothing }\to {\mathbb{P}}_S^0$ is a closed immersion by the same argument as step 1.3, and its composite with the isomorphism ${\mathbb{P}}_S^0\to S$ is the empty morphism. Thus $\mathrm{\varnothing }\to S$ factors as a closed immersion into ${\mathbb{P}}_S^0$ followed by the projection, so it is projective by [F9]. The argument is choice-free, and the case $S=\mathrm{\varnothing }$ is included: then the empty morphism is the identity of the empty scheme, which steps 1.1-2.1 still treat through $0=\mathrm{Spec}0$.