Let $$p:X\rightarrow Y$$ be a closed, continuous, surjective map. I need to show that if $X$ is normal, then $Y$ is also normal.
So I used this result: A space $X$ is normal iff for any closed set $C$ in $X$, and any nbd $U$ containing $C$, there exists a nbd $W$ of $C$ such that $$C\subset \bar W \subset U$$
Now, take any closed set $B$ of $Y$ and $V$ be and open set in $Y$ containing $B$. $$B\subset V\subset Y$$ Define $A=p^{-1}(B)$ and $U=p^{-1}(V)$. By continuity of $p$, $A$ is closed in $X$, $U$ is open in $X$. Also $$B\subset V\\i.e.\ \ p^{-1}(B)\subset p^{-1}(V)\\i.e.\ \ A\subset U.$$
Since $X$ is normal, there exists an open set $W$ in $X$ such that $$A\subset \bar W\subset U.$$ As $p$ is closed and surjective, from the last step we can write $$p(A)\subset p(\bar W) \subset p(U).$$ Thus the statement is proved. Is this correct?
Also there is was hint given: If $U$ is any open set containing $p^{-1}(\{y\})$ then there is an open set $W$ in $Y$ containing $y$ such that $p^{-1}(W)\subset U$
Although I have managed to prove the statement in the hint, I could not figure out how to use that in the proof.
Somehow I know my version of the proof must be wrong.
Please help.