Proof verification on Munkres Chapter 4, Section 31, Problem 7 Let $p: X \to Y$ be a closed surjective continuous map such that $p^{-1} ( \{ y \} )$ is compact for every $y \in Y$. Show that if $X$ is regular, then so is $Y$. (In this book, regular means $T_3$; I found some books that defines $T_3$ axiom as $T_1$ and regular.)

Lemma If $U$ is an open set containing $p^{-1} ( \{ y \} )$, then there exists a neighborhood $W$ of $y$ such that $p^{-1} (W) \subset U$.

For any closed set $B \subset Y$ and point $y \notin B$, let $A = p^{-1} (B)$ and $W = p^{-1} ( \{ y \} )$. By assumption $A$ is closed and $W$ is compact. For every $x \in W$, there exists a neighborhood $U_x$ of $x$ and $V_x$ of $A$ which is disjoint. The collection $\{ U_x : x \in W \}$ is the open cover of $W$. Since $W$ is compact, there exists a finite set $S \subset W$ such that $P = \bigcup_{x \in S} U_x$ covers $W$. It is clear that $Q = \bigcap_{x \in S} V_x$ is a neighborhood of $A$.
By the lemma, there exists a neighborhood $M$ of $y$ such that $p^{-1} (M) \subset P$, and for every point $z \in B$, there exists a neighborhood $W_z$ of $z$ such that $p^{-1} (W_z) \subset Q$. $\bigcup_{z \in B} W_z$ and $N$ is disjoint neighborhood of $B$ and $y$, respectively.

Does my proof work without any critical gaps?
Since I am not a native English user, advice on my language skills will also be appreciated.
 A: Yes, that classic lemma is the standard way to go:
The following more general lemma is true:

if $p$ is closed and onto, let $O$ be an open neighbourhood of $p^{-1}[A]$ for some $A \subseteq Y$, then there is an open subset $O' \subseteq Y$ so that $A \subseteq O'$ and $p^{-1}[O']\subseteq O$.

Your lemma is only stated for $A=\{y\}$. But the proof is the same in either case.
So $y \notin C$ where $C$ is closed, then $B:=p^{-1}[C]$ is closed and $F_y:= p^{-1}[\{y\}]$ is compact.
The standard separation axiom: point to compact set trick applies;
For each $x \in F_y$ we have open $U_x$ and $V_x$ so that $U_x \cap V_x = \emptyset$ and $x \in U_x; B\subseteq V_x$. Let $\{U_x: x \in N\}$ ($N$ finite) be a finite subcover of $F_y$ and define $U=\bigcup_{x \in N} U_x \supseteq F_y$ and $V:= \bigcap_{x \in F} V_x\supseteq B$ (intersection! not union as you did) and $U \cap V = \emptyset$.
The generalised lemma then gives us $U' \ni y$ and $V'\supseteq C$ open in $Y$ so that $p^{-1}[U'] \subseteq U$ and $p^{-1}[V'] \subseteq V$ and as $p$ is onto $U'$ and $V'$ are disjoint, because $U$ and $V$ are, and we have regularity..
