Here is a nice example. Let $X$ be a set with $|X|>2$, and give $X$ the trivial topology (so the only open sets are $\varnothing$ and $X$). Let $a,b\in X$ be distinct points. Then the map $f:X\to X$ with $f(a)=b$, $f(b)=a$, and $f(x)=x$ for any $x\neq a,b$ has as its set of fixed points $X\setminus\{a,b\}$, which is not closed because the only closed sets are $\varnothing$ and $X$.
It is easy to see that $X$ is not Hausdorff; because the only open sets available to us are $\varnothing$ and $X$, we can't separate distinct points with disjoint open sets.
For the hint you have been given, you have given the correct set $W$.
Note that as $p^{-1} [\{ y \}] \subseteq U$, then $p(x) \neq y$ for all $x \in X \setminus U$, which implies that $y \notin p [ X \setminus U ]$, or, equivalently, $y \in Y \setminus p [ X \setminus U ] = W$.
I would be tempted to attack this problem in a slightly different manner, noting that essentially by de Morgan's Laws, normality of a topological space $X$ is equivalent to the following:
Given open $U , V \subseteq X$ such that $U \cup V = X$ there are closed $E \subseteq U$ and $F \subseteq V$ such that $E \cup F = X$.
So let $U,V \subseteq Y$ be open sets such that $U \cup V = Y$. Then by continuity of $f$, $f^{-1}[U], f^{-1}[V]$ are open subsets of $X$, and $f^{-1}[U] \cup f^{-1}[V] = X$. As $X$ is normal the condition above implies that there are closed $E \subseteq f^{-1}[U]$ and $F \subseteq f^{-1}[V]$ such that $E \cup F = X$. It is easy to check that $f[E] \subseteq U$ and $f[F] \subseteq V$. As $f$ is a closed mapping, then $f[E],f[F]$ are closed subsets of $Y$, and by the surjectivity of $f$ it follows that $f[E] \cup f[F] = Y$. Thus $f[E],f[F]$ are as required.
Best Answer
Let $X$ be $\Bbb N$ with the discrete topology, let $Y$ be $\Bbb N$ with the cofinite topology, and let $f:X\to Y$ be the identity map. Clearly $X$ is normal and $f$ is continuous and surjective. However, $Y$ is $T_1$ but not Hausdorff, so it clearly is not normal.