Let $(X,\mathscr T)$ be normal space, let $(Y,\mathscr U)$ be topological spaces, and let $f:X\to Y$ be closed continuous function that maps $X$ onto $Y$. Then $(Y,\mathscr U)$ is normal.
Proof. Let $C$ and $D$ be disjoint closed subsets of $(Y,\mathscr U)$. Then $f^{-1}(C)$ and $f^{-1}(D)$ are disjoint closed subsets $(X,\mathscr T).$ By Normality, there exists disjoint open sets $U$ and $V$ in $(X,\mathscr T)$ such that $f^{-1}(C)\subseteq U$ and $f^{-1}(D)\subseteq V$. Since $f$ is closed, $f(X\setminus U)$ and $f(X\setminus V)$ are closed in $(Y,\mathscr U)$. Therefore $M=Y\setminus f(X\setminus U)$ and $N=Y\setminus f(X\setminus V)$ are open. But I could prove $C\subseteq M, D\subseteq N,$ and $M\cap N=\emptyset$ only by assuming $f$ is injective. But author didn't specify the injectivity of the function. Was he mistaken?
Suppose $C \subseteq M$ failed. Then we'd have $c \in C$ with $c \notin M$. But $c \notin M$ implies $c \in f[X\setminus U]$ by the definition of $M$. So $c = f(x)$ with $x \in X\setminus U$. Then $x \in f^{-1}[C]$ (its image is in $C$) but $x \notin U$ contradicting $f^{-1}[C]\subseteq U$. This contradiction shows the inclusion does hold.
$D \subseteq N$ is entirely similar.
Suppose $y \in M \cap N$. As $f$ is surjective (we do need that here) $y=f(x)$ for some $x \in X$. Then $x \notin X\setminus U$ (or else $y=f(x) \in f[X\setminus U]$ which is disjoint from $M$) so $x \in U$. Similarly $x \in V$. Contradiction, as $U$ and $V$ were chosen disjoint. Hence $M$ and $N$ are disjoint.