$X$ normal $f:X \longrightarrow Y$ continuous, surjective, closed $\Longrightarrow$ $Y$ normal

Let $y\in E$, and suppose that $y\notin U'$. Then $y\in Y\setminus U'=f[X\setminus U]$, so there is an $x\in X\setminus U$ such that $f(x)=y$. But then $x\in f^{-1}[E]\subseteq U$, which is a contradiction.

I suspect that what you were missing is the fact that $f^{-1}[E]$ contains every point of $X$ that maps to $E$.

For every $x \notin U$ we have in particular $x \notin f^{-1}(E)$ and therefore $f(x) \notin E$. So $f(U^c) \cap E = \emptyset$ and therefore $E \subseteq f(U^c)^c = U'$. Similarly $V \subseteq f(V^c)^c$.

Notice that we need the surjectivity to conclude that $U' \cap V' = \emptyset$.