Prob. 7 (b), Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The image of a regular space under a perfect map is also a regular space Here is Prob. 7 (b), Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Let $p \colon X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1} \big( \{ y \} \big)$ is compact for each $y \in Y$. (Such a map is called a perfect map.) Show that if $X$ is regular, then so is $Y$.

My Attempt:

As topological space $X$ is a regular space, so $X$ is also a Hausdorff space, and as $p \colon X \rightarrow Y$ is a perfect map, so $Y$ is also a Hausdorff space, by Prob. 7 (a), Sec. 31, in Munkres' Topology, 2nd edition.

Here is my Math Stack Exchange post on Prob. 1, Sec. 31, in Munkres' Topology, 2nd edition.

Now as $Y$ is a Huasdorff space, so one-point sets in $Y$ are closed, by Theorem 17.8 in Munkres.
Let $y$ be a point of $Y$, and let $B$ be any closed set of $Y$ such that $y \not\in B$. We need to find disjoint open sets $U$ and $V$ of $Y$ such that $y \in U$ and $B \subset V$. 
Now as $B$ is a closed set in $Y$ and as the map $p \colon X \rightarrow Y$ is continuous, so the inverse image set $p^{-1}(B)$ is a closed set of $X$, by Theorem 18.1 (3) in Munkres' Topology, 2nd edition.
As $p \colon X \rightarrow Y$ is a surjective map and as $y$ is a point of $Y$, so there exists a point $x$ of $X$ for which $y = p(x)$, and as $y = p(x) \not\in B$, so the point $x \not\in p^{-1}(B)$, where $x$ is any point of $X$ such that $p(x) = y$; let us pick one such point $x$ arbitrarily; there may be more than one such $x$ depending on the (lack of) injectivity of the mapping $p \colon X \rightarrow Y$; then as that point  $x$ is a point of the regular space $X$ and $p^{-1}(B)$ is a closed set in $X$ such that $x \not\in p^{-1}(B)$, so there exist disjoint open sets $U_x^\prime$ and $V_x^\prime$ of $X$ such that 
  $$ x \in U_x^\prime \qquad \mbox{ and } \qquad  p^{-1}(B) \subset V_x^\prime. \tag{0} $$
Thus the collection 
  $$ \left\{ \, U_x \, \colon \, x \in X, p(x) = y \, \right\} $$
  is a covering of $p^{-1}\big( \{ y \} \big)$ by sets open in $X$, and as $p^{-1}\big( \{ y \} \big)$ is a compact subspace of $X$ by our hypothesis, so we can conclude that this collection has a finite subcollection 
  $$
\left\{ \, U_{x_1}, \ldots, U_{x_n} \, \right\}
$$
  also covering $p^{-1}\big( \{ y \} \big)$, where $x_1, \ldots, x_n \in X$ such that 
  $$
y = p \left( x_1 \right) = \cdots = p \left( x_n \right). 
$$
  Let us put 
  $$
U^\prime \colon= \bigcup_{i=1}^n U_{x_i} \qquad \mbox{ and } \qquad V^\prime \colon= \bigcap_{i=1}^n V_{x_i}. \tag{Definition 0}
$$
  Then $U^\prime$ and $V^\prime$ are open sets of $X$ such that 
  $$
p^{-1}\big( \{ y \} \big) \subset U^\prime \qquad \mbox{ and } \qquad p^{-1}(B) \subset V^\prime. \tag{1}
$$
  Please refer to (0) and (Definition 0) above.
Now we show that $U^\prime$ and $V^\prime$ are disjoint. Let $x \in V^\prime$. Then $x \in V_{x_i}^\prime$ for each $i = 1, \ldots, n$ [Refer to (Definition 0) above.], and then $x \not\in U_{x_i}$ for each $i = 1, \ldots, n$, which implies that $x \not\in U^\prime$. Thus $U^\prime$ and $V^\prime$ are indeed disjoint open sets of $X$. 
Now as $U^\prime$ and $V^\prime$ are open sets in $X$, so $X \setminus U^\prime$ and $X \setminus V^\prime$ are closed sets of $X$, and as the map $p \colon X \rightarrow Y$ is a closed map by our hypothesis, so the sets $p \left( X \setminus U^\prime \right)$ and $p \left( X \setminus V^\prime \right)$ are closed sets in $Y$, and therefore the sets $Y \setminus p \left( X \setminus U^\prime \right)$ and $Y \setminus p \left( X \setminus V^\prime \right)$ are open in $Y$. 
We now show that the sets $Y \setminus p \left( X \setminus U^\prime \right)$ and $Y \setminus p \left( X \setminus V^\prime \right)$ are disjoint, we suppose this is not the case, and let $t \in \left( Y \setminus p \left( X \setminus U^\prime \right) \right) \cap \left( Y \setminus p \left( X \setminus V^\prime \right) \right)$. Then $t \in Y \setminus p \left( X \setminus U^\prime \right)$ and $t \in Y \setminus p \left( X \setminus V^\prime \right)$, which implies that $t \in Y$ such that $t \not\in p \left( X \setminus U^\prime \right)$ and $t \not\in p \left( X \setminus V^\prime \right)$, and as the mapping $p \colon X \rightarrow Y$ is surjective, so we can conclude that there exists a point $s \in X$ such that $t = p(s)$ and any such point $s$ also satisfies $s \not\in X \setminus U^\prime$ and $s \not\in X \setminus V^\prime$, and therefore $s \in U^\prime$ and $s \in V^\prime$, which contradicts the fact that $U^\prime$ and $V^\prime$ are disjoint. Thus we can conclude that the sets $Y \setminus p \left( X \setminus U^\prime \right)$ and $Y \setminus p \left( X \setminus V^\prime \right)$ are disjoint open sets in $Y$.
From (1) above, we obtain
  $$
X \setminus U^\prime \subset X \setminus p^{-1} \big( \{ y \} \big),
$$
  which implies
  $$
p \left( X \setminus U^\prime \right) \subset p \left( X \setminus p^{-1} \big( \{ y \} \big) \right),
$$
  and hence
  $$
Y \setminus p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) \subset Y \setminus p \left( X \setminus U^\prime \right). \tag{2}
$$
  But as $p \colon X \rightarrow Y$ is a surjective mapping, so we also get
  $$
\begin{align}
Y \setminus p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) &= Y \setminus p \left( p^{-1}(Y) \setminus p^{-1} \big( \{ y \} \big) \right) \\
&= Y \setminus p \left( p^{-1} \big( Y \setminus \{ y \} \big) \right) \\
&= Y \setminus \big( Y \setminus \{ y \} \big) \\
&= \{ y \},
\end{align}
$$
  that is,
  $$
Y \setminus p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) = \{ y \}. \tag{3} 
$$
  From (2) and (3) we can conclude that
  $$
\{ y \} \subset Y \setminus p \left( X \setminus U^\prime \right),
$$
  that is,
  $$
y \in Y \setminus p \left( X \setminus U^\prime \right). \tag{4} 
$$
Also from (1) above, we have 
  $$
X \setminus V^\prime \subset X \setminus p^{-1}(B),
$$
  which implies
  $$
p \left( X \setminus V^\prime \right) \subset p \left( X \setminus p^{-1}(B) \right),
$$
  and hence 
  $$
Y \setminus p \left( X \setminus p^{-1}(B) \right) \subset  Y \setminus p \left( X \setminus V^\prime \right). \tag{5} 
$$
  And again using the surjectivity of the map $p \colon X \rightarrow Y$, we find that 
  $$
\begin{align}
Y \setminus p \left( X \setminus p^{-1}(B) \right) &= Y \setminus p \left( p^{-1}(Y) \setminus p^{-1}(B) \right) \\
&= Y \setminus p \left( p^{-1} \big( Y \setminus B \big) \right) \\
&= Y \setminus \left( Y \setminus B \right) \\
&= B,
\end{align}
$$
  that is,
  $$
Y \setminus p \left( X \setminus p^{-1}(B) \right)  = B,
$$
  which together with (5) above gives
  $$
B \subset Y \setminus p \left( X \setminus V^\prime \right). \tag{6}
$$
Thus we have shown that 
(1) one-point sets are closed in $Y$, and 
(2) for any point $y \in Y$ and for any closed set $B$ of $Y$ not containing the point $y$, there exist disjoint open sets $U \colon= Y \setminus p \left( X \setminus U^\prime \right)$ and $V \colon= Y \setminus p \left( X \setminus V^\prime \right)$ of $Y$ such that $y \in U$ and $B \subset V$. Refer to (4) and (6) above. 
Hence $Y$ is a regular space.

Is my proof good enough? Or, are there problems?
 A: The fact that $Y$ is $T_1$ (i.e. singletons are closed) is immediate from two facts: if $y \in Y$, $p^{-1}[\{y\}]$ is closed (it's compact in a Hausdorff space!) and $p$ is a quotient map (from being closed, continuous and onto); or you conclude it right away from Hausdorffness of $X$ and the earlier problem on that so that $Y$ is Hausdorff and hence $T_1$, which is the most efficient.
The lemma I quoted in my earlier post on the Hausdorff version of this problem, can be generalised (without a lot of effort; you're sort of reproving it too in the final part, though with more set computation instead of point reasoning (as I did in my proof linked to in my link..)) as

A function $f: X \to Y$ between topological spaces is closed iff for every $B \subseteq Y$ and every open $U$ such that $f^{-1}[B]\subseteq U$ there exists an open $V$ in $Y$ such that $B \subseteq V$ and $f^{-1}[V]\subseteq U$.

and assuming that we just have to remark that in a regular space we can separate a  compact set $K$ and a closed set $C$ that are disjoint by disjoint open sets (same proof as points/compact sets in Hausdorff spaces) and apply the same idea:
If $y \notin C$ and $C$ closed in $Y$, $p^{-1}[\{y\}$ is compact and disjoint from $p^{-1}[C]$ (which is closed by continuity). So as $X$ is regular, we have disjoint open sets $U$ and $U'$ in $X$ such that $p^{-1}[\{y\}] \subseteq U$ and $p^{-1}[C]\subseteq U'$. Apply the lemma for closed maps twice to get $V$ open with $y \in V$ and $V'$ open with $C \subseteq V'$ so that $p^{-1}[V] \subseteq U$ and $p^{-1}[V'] \subseteq U'$, and then surjectivity of $p$ implies that $V \cap V'=\emptyset$ and so $Y$ is regular.
This lemma-use makes the proof much more convenient to follow and saves a lot of time writing up the essentially same arguments in several posts. It will also be useful in other proofs involving closed maps, so it's a good tool to have. Engelking in his excellent General Topology has also isolated as a separate proposition to re-use several times. This is where I first saw it too. 
