Prop. If $f\colon X\longrightarrow Y$ is a separated morphism of schemes, then every section $\sigma\colon Y\longrightarrow X$ is a closed immersion.

Proof. Let $\sigma\in X(Y)$ be a section of the $Y$-scheme $X$. Let's take the fibered product $\left(X\times_X Y,p_X,q_X\right)$, where we are considering $Y$ as a $X$-scheme endowing it with the structural morphism $\sigma\colon Y\longrightarrow X$. The projections $p_X$ and $q_X$ induce a closed immersion of $Y$-schemes $X\times_X Y\longrightarrow X\times_Y Y$, since $\ast$ $f\colon X\longrightarrow Y$ is separated. In a similar way, the identity map of $Y$ and the section $\sigma$ induce a morphism $Y\longrightarrow X\times_X X$ of $X$-schemes. In other words we have the following commutative diagram.

At this point the solution of the exercise follows by the remark about the composition of closed immersions in the beginning of the solution and by the fact that the morphisms $q^{-1}_X$ and $p^{}_Y$ are isomorphisms, then, in particular, closed immersions.

$\ast$ Let $Y$ be a separated $Z$-scheme. Then for any $Y$-schemes $X_1,X_2$, the canonical morphism $X_1\times_Y X_2\longrightarrow X_1\times_Z X_2$ is a closed immersion.

The claim above is the Proposition 3.9 (f) at pag.101 of Algebraic Geometry and Aritmetic Curves.

Prop. If $f\colon X\longrightarrow Y$ is a separated morphism of schemes, then every section $\sigma\colon Y\longrightarrow X$ is a closed immersion.

Proof. Let $\sigma\in X(Y)$ be a section of the $Y$-scheme $X$. Let's take the fibered product $\left(X\times_X Y,p_X,q_X\right)$, where we are considering $Y$ as a $X$-scheme endowing it with the structural morphism $\sigma\colon Y\longrightarrow X$. The projections $p_X$ and $q_X$ induce a closed immersion of $Y$-schemes $X\times_X Y\longrightarrow X\times_Y Y$, since $\ast$ $f\colon X\longrightarrow Y$ is separated. In a similar way, the identity map of $Y$ and the section $\sigma$ induce a morphism $Y\longrightarrow X\times_X X$ of $X$-schemes. In other words we have the following commutative diagram.

At this point the solution of the exercise follows by the fact that a composition of closed immersions is a closed immersion and by the fact that the morphisms $q^{-1}_X$ and $p^{}_Y$ are isomorphisms, then, in particular, closed immersions.

$\ast$ Let $Y$ be a separated $Z$-scheme. Then for any $Y$-schemes $X_1,X_2$, the canonical morphism $X_1\times_Y X_2\longrightarrow X_1\times_Z X_2$ is a closed immersion.

The claim above is the Proposition 3.9 (f) at pag.101 of Algebraic Geometry and Aritmetic Curves.