Image of a normal space under a closed and continuous map is normal 
$p : X \to Y$ is continuous, closed and surjective, and $X$ is a normal space. Show $Y$ is normal.

There is a hint, which I'm trying to prove: show that if $U$ is open in $X$ and $p^{-1}(\{y\}) \subset U$, $y \in Y$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W) \subset U$.
I have a candidate for $W$, namely $W=Y\setminus p(X \setminus U)$. I did prove that this $W$ is open, and that $p^{-1}(W) \subset U$, but I don't see how $y \in W$. I think this would require injectivity of $p$.
I have also shown that $y \in p(U)$ and that $W \subset p(U)$, so if also $W \supset p(U)$, then $y \in W$. 
Can anyone help me?
 A: Why go through all that hassle, when you can do as follows:  
Let $p: X \rightarrow Y$ be a closed, continuous surjection. Now let $A,B$ be two disjoint closed subsets of $Y$. Because $X$ is normal, we can separate the closed disjoint sets $p^{-1}(A), p^{-1}(B)$ in $X$ by respective neighborhoods $U_1, U_2$. Now choose neighborhoods $V_1$ of $A$, and $V_2$ of $B$ s.t. $p^{-1}(V_1) \subset U_1$, and $p^{-1}(V_2) \subset U_2$. Then it follows that $V_1, V_2$ are disjoint. Hence, $Y$ is normal.  
Note that in general, a continuous image of a normal space is not necessarily normal.
A: 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.
