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?
