This is related to Ueno's Algebraic Geometry 2, Lemma 5.8 proof.
Let $f:X\to Y$ be quasicompact morphism of schemes (i.e. there is covering $U_i$ of $Y$ s.t. $f^{-1}(U_i)$ are quasi-compact). Then $f(X)$ is closed iff $f(X)$ is closed under specialization(i.e. if $x\in f(X)$ and $y\in\overline{x}$, then $y\in f(X)$).
Assume $f(X)$ is closed under specialization. Since reduced schemes have same topology as before, it suffices to assume both $X,Y$ are reduced. Replace $Y$ by $\overline{f(X)}$. If $y\in Y$, we have to check $y\in f(X)$. It suffices to assume $Y=Spec(A)$ by refining covering via quasi-compactness of $f$. From quasi-compactness of $f$, one has $f^{-1}(Y)$ is covered by finite many affines $\operatorname{Spec}(B_i)$of $X$. It suffices to assume $i=1$. Hence $f$ is given by $f:A\to B_1$.
$\textbf{Q:}$ The book says one can replace $A$ by $\operatorname{Im}(A)$ to get $f$ injection. Why can I do so? Consider $0\to \operatorname{Ker}(f)\to A\to \operatorname{Im}(A)\to 0$. Now the goal is to check $f$ is epimorphism. This at least requires $\operatorname{Ker}(f)$ nilradical as all prime ideals of $A$ have to be seen by $A\to B_1$ ring morphism.