A morphism $f : X \to Y$ is a closed immersion if and only if there exists an affine open cover $ \{ U_i \}$ such that $f|_{f^{-1}(U_i)} : f^{-1}(U_i) \to U_i $ is a closed immersion for all $i$.
It's so fundamental, but I can't show it. The stalk property - for all $ x \in X$ $f^{\#}_x : \mathscr{O}_{Y,f(x)} \to \mathscr{O}_{X,x} $ is surjective - is trivial, but I don't understand why $f(X)$ is closed in $Y$.
$f(X)$ is closed:
Let's show instead $Y\setminus f(X)$ is open. First, since each of the maps $f^{-1}(U_i)=U_i\cap f(X)\rightarrow U_i$ is closed, we can write $$U_i\setminus (f(X)\cap U_i)=\bigcup_jD(f_{ij})$$ for some elements $f_{ij}$ in $\mathcal{O}_{U_i}(U_i)$. Now, doing this for all of the $U_i$ allows us to write $$Y\setminus f(X)=\bigcup_i U_i\setminus (f(X)\cap U_i)= \bigcup_{i,j} D(f_{i,j})$$ which proves the claim.
