The Stacks project

Proof. This is clear for the case of open immersions since an open subspace of an open subspace is also an open subspace.

Suppose $a : Z \to Y$ and $b : Y \to X$ are closed immersions of schemes. We will verify that $c = b \circ a$ is also a closed immersion. The assumption implies that $a$ and $b$ are homeomorphisms onto closed subsets, and hence also $c = b \circ a$ is a homeomorphism onto a closed subset. Moreover, the map $\mathcal{O}_ X \to c_*\mathcal{O}_ Z$ is surjective since it factors as the composition of the surjective maps $\mathcal{O}_ X \to b_*\mathcal{O}_ Y$ and $b_*\mathcal{O}_ Y \to b_*a_*\mathcal{O}_ Z$ (surjective as $b_*$ is exact, see Modules, Lemma 17.6.1). Hence by Lemma 26.24.2 above $c$ is a closed immersion.

Finally, we come to the case of immersions. Suppose $a : Z \to Y$ and $b : Y \to X$ are immersions of schemes. This means there exist open subschemes $V \subset Y$ and $U \subset X$ such that $a(Z) \subset V$, $b(Y) \subset U$ and $a : Z \to V$ and $b : Y \to U$ are closed immersions. Since the topology on $Y$ is induced from the topology on $U$ we can find an open $U' \subset U$ such that $V = b^{-1}(U')$. Then we see that $Z \to V = b^{-1}(U') \to U'$ is a composition of closed immersions and hence a closed immersion. This proves that $Z \to X$ is an immersion and we win. $\square$