Prob. 7 (a), Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The image of a Hausdorff space under a perfect map is also a Hausdorff space Here is Prob. 7 (a), 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 Hausdorff, then so is $Y$.

My Attempt:

Let $u$ and $v$ be any two distinct points of $Y$. Then the inverse image sets $p^{-1}\left( \left\{ u \right\} \right)$ and $p^{-1}\left( \left\{ v \right\} \right)$ are non-empty (because $p$ is onto), disjoint (because the sets $\left\{ u \right\}$ and $\left\{ v \right\}$ are disjoint), and compact (by our hypothesis about the map $p$) subsets of $X$. 
Now as $X$ is a Hausdorff space and as  $p^{-1}\left( \left\{ u \right\} \right)$ and $p^{-1}\left( \left\{ v \right\} \right)$ are disjoint compact subspaces of $X$, so by Prob. 5, Sec. 26, in Munkres there exist disjoint open sets $U^\prime$ and $V^\prime$ of $X$ containing $p^{-1}\left( \left\{ u \right\} \right)$ and $p^{-1}\left( \left\{ v \right\} \right)$, respectively. 

Here is my Math Stack Exchange post on Prob. 5, Sec. 26, in Munkres' Topology, 2nd edition.

Now as $U^\prime$ and $V^\prime$ are open sets in $X$, so the sets $X \setminus U^\prime$ and $X \setminus V^\prime$ are closed, and as $p \colon X \rightarrow Y$ is a closed map, so the image sets $p\left( X \setminus U^\prime \right)$ and $p \left( X \setminus V^\prime \right)$ are closed in $Y$, and thus 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; suppose if possible that these sets are not disjoint.
Let 
  $$ y \in \left( Y \setminus p\left( X \setminus U^\prime \right) \right) \cap \left( Y \setminus p \left( X \setminus V^\prime \right) \right). $$ 
  Then $y \in Y \setminus p\left( X \setminus U^\prime \right)$ and  $y \in Y \setminus p\left( X \setminus V^\prime \right)$. So $y \in Y$ such that $y \not\in p \left( X \setminus U^\prime \right)$ and $y \not\in p \left( X \setminus V^\prime \right)$, and as the map $p \colon X \rightarrow Y$ is a surjective map, so we can conclude that there exists a point $x \in X$ for which $y = p(x)$ and that point $x \not\in X \setminus U^\prime$ and $x \not\in X \setminus V^\prime$, which implies that $x \in U^\prime$ and $x \in V^\prime$, and hence $x \in U^\prime \cap V^\prime$, which contradicts our choice of $U^\prime$ and $V^\prime$ being disjoint. Please refer to the second paragraph of this proof. Therefore 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 two disjoint open sets in $Y$. Plese refer to the preceding paragraph.
Now as 
  $$ p^{-1} \left( \left\{ u \right\} \right) \subset U^\prime, $$
  so we can conclude that 
  $$
X \setminus U^\prime \subset X \setminus p^{-1} \left( \left\{ u \right\} \right),
$$
  which implies that
  $$
p \left( X \setminus U^\prime \right) \subset p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right),
$$
  and hence
  $$
Y \setminus p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right) \subset Y \setminus p \left( X \setminus U^\prime \right). \tag{1}
$$
  And similarly, we also obtain 
  $$
Y \setminus p \left( X \setminus p^{-1} \left( \left\{ v \right\} \right) \right) \subset Y \setminus p \left( X \setminus V^\prime \right). \tag{2}
$$
Now as $p \colon X \rightarrow Y$ is a surjective map and as $u \in Y$, so we can conclude that there exists a point $x \in X$ for which $u = p \left( x  \right)$, and any such point $x$ satisfies $x \in p^{-1} \left( \left\{ u \right\} \right)$, and then any such $x  \not\in X \setminus p^{-1} \left( \left\{ u \right\} \right)$, which implies that $u = p \left( x  \right) \not\in p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right)$, and therefore 
  $u \in Y \setminus p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right)$, which by virtue of (1) above implies that $u \in Y \setminus p \left( X \setminus U^\prime \right)$. 
And, by analogous reasoning we can conclude from (2) above that $v \in Y \setminus p \left( X \setminus V^\prime \right)$. 
Thus we have shown that, given any two distinct points $u$ and $v$ of $Y$, there exist two disjoint open sets $U \colon= Y \setminus p \left( X \setminus U^\prime \right)$ and $V \colon= Y \setminus p \left( X \setminus V^\prime \right)$ containing $u$ and $v$, respectively.
Hence $Y$ is a Hausdorff space.

PS: 
Having obtained (1) and (2) above, we can also proceed as follows:

As $p \colon X \rightarrow Y$ is a surjective mapping, so we find that
  $$
\begin{align}
Y \setminus p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right) &= Y \setminus p \left( p^{-1}(Y) \setminus p^{-1} \big( \{ u \} \big) \right) \\
&= Y \setminus p \left( p^{-1} \big( Y \setminus \{ u \} \big) \right) \\
&= Y \setminus \big( Y \setminus \{ u \} \big) \\
&= \{ u \},
\end{align}
$$
  that is, 
  $$
Y \setminus p \left( X \setminus p^{-1} \left( \left\{ u \right\} \right) \right) = \{ u \},
$$
  and then (1) gives 
  $$
\{ u \} \subset Y \setminus p \left( X \setminus U^\prime \right),
$$
  that is,
  $$
u \in Y \setminus p \left( X \setminus U^\prime \right).
$$
  And similarly, we also obtain
  $$
v \in Y \setminus p \left( X \setminus V^\prime \right).
$$

Is my proof correct and spelled out clearly enough? Or, are there issues of accuracy or clarity in my attempt?
 A: It is correct but I think the last part could be shortened. We want to show $u \in U.$ If not, then $u \in p(X\setminus U').$ Therefore there exists $x \in X \setminus U'$ such that $u=p(x).$ Thus $x \in p^{-1}(\{u\})\subseteq U',$ which is a contradiction. So $u \in U$ and similarly $v \in V.$
A: Introduce the following lemma:

A function $p: X \rightarrow Y$ between topological spaces $X$ and $Y$ is a closed map if and only if, for every point $y \in Y$ and for every open set $U$ in $X$ such that $p^{-1} \big[ \{ y \} \big] \subseteq U$, there exists an open set $V$ in $Y$ such that $y \in V$ and $p^{-1}[V]\subseteq U$.

(A sort of reverse continuity wrt fibres; I showed it here e.g., it also demonstrates how to shorten your own proof, because you essentially use one direction of it).
Then if $y \neq y'$ the fibres $p^{-1}[\{y\}]$ and $p^{-1}[\{y'\}]$ are disjoint, compact so in a Hausdorff space they have disjoint neighbourhoods $U$ resp. $U'$. The promised $V$ and $V'$ from the lemma for $U$ resp $U'$ are then also disjoint by surjectivity of $p$ (in that case we can conclude from the disjointness of $p^{-1}[V]$ and $p^{-1}[V']$ the disjointness of $V$ and $V'$).
So it's a combination of the above lemma plus the second lemma you quote as Prob 5, sec. 26, that in a Hausdorff space we can separate not only points, but also disjoint compact sets. It's conceptually easier to split it up that way, I think. It also makes clearer that continuity of $f$ is irrelevant for this result, just closedness plus compact fibres (and ontoness) are used.
