The image of a normal space under a closed map is normal 
Let $X$ be normal and $p: X \to X/R$ be a closed and open map. Show that $X/R$ is normal.

First, I don't understand how can it be closed and open at the same time, I'm confused.
I search and a space $X$ is normal if for any $A, B$ closed in $X$ there is an open set $U$ for $A$ and an open set $V$ for $B$ such that $U \cap V = \emptyset$, is that right? So, I have to find two open sets which are disjoint in $X/R$?
 A: Suppose $f: X \to Y$ is closed and continuous and onto and $X$ is normal. Take $C$ and $D$ disjoint and closed in $Y$, then $f^{-1}[C]$ and $f^{-1}[D]$ are disjoint and closed in $X$ so have disjoint open neighbourhoods $U$ and $V$ so that
$$U \cap V = \emptyset, f^{-1}[C] \subseteq U, f^{-1}[D] \subseteq V$$
Define $O_1 = Y\setminus f[X\setminus U]$, $O_2 =  Y\setminus f[X\setminus V]$ which are both open as $f$ is a closed map. ($U$ open so $X\setminus U$ closed, so its image under $f$ closed in $Y$ so its complement in $Y$ open there).
Suppose $y \in O_1 \cap O_2$. Then $y = f(x)$ for some $x$. Then $x \notin X\setminus U$ (or otherwise $y \in f[X\setminus U]$, so $y \notin O_1$, contradiction), hence $x \in U$. By the same reasoning $x \in V$ contradicting $U \cap V = \emptyset$, so no such $y$ can exist and $O_1 \cap O_2$.
Now $C \subseteq O_1$ and $D \subseteq O_2$ are also clear: Suppose e.g. $y \in C$, and let $x \in X$ be such that $f(x) = y$. Then $x \in f^{-1}[C] \subseteq U$, so $x \in U$ and as this holds for all such $x$, $y \notin f[X\setminus U]$, so $y \in O_1$, etc.
Hence $Y$ is normal.
We don't need $f$ is open, just continuous and closed.
