Prob. 6, Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The image of a normal space under a closed continuous map is also normal Here is Prob. 6, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Let $p \colon X \to Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$. [Hint: If $U$ is an open set containing $p^{-1}(\{ y \} )$, show there is a neighborhood $W$ of $y$ such that $p^{-1}(W) \subset U$.]

My Attempt:

For any  point $y\in Y$, we can find a point $x \in X$ such that $y = p(x)$, because $p$ is surjective. As $X$ is normal, so the set $\{ x \}$ is a closed set in $X$, and since $p$ is a closed map, the set $p(\{ x \}) = \{p(x) \} = \{y\}$ is a closed set in $Y$. Thus one-point sets are closed in $Y$. 
Let $B$ be a closed set in $Y$, and let $U$ be an open set in $Y$ containing $B$. We need to find an open set $V$ in $Y$ such that $B \subset V$ and $\overline{V} \subset U$, by Lemma 31.1 (b) in Munkres.
As $p \colon X \to Y$ is continuous and as sets $B$ and $U$ are, respectively, closed and open in $Y$, so the inverse images $p^{-1}(B)$ and  $p^{-1} (U)$ are, respectively, closed and open in $X$. 
And, as $B \subset U$, so we must also have 
  $$ p^{-1}(B) \subset p^{-1}(U). \tag{0} $$
Thus $p^{-1}(B)$ is a closed set in $X$ and $p^{-1}(U)$ is an open set in $X$ such that (0) holds. So using Lemma 31.1 (b) in Munkres and normality of $X$, we can conclude that there exists an open set $W$ in $X$ such that 
  $$ p^{-1}(B) \subset W \qquad \mbox{ and  } \qquad \overline{W} \subset p^{-1}(U). \tag{A} $$
As $p \colon X \to Y$ is continuous, so we must have 
  $$ p\left( \overline{W} \right) \subset \overline{ p(W) }, \tag{1} $$
  by Theorem 18.1 (2) in Munkres. 
On the other hand, as  $W \subset \overline{W}$, so 
  $$ p(W) \subset p \left( \overline{W} \right). $$
  Moreover, as $p \colon X \to Y$ is a closed map and as $\overline{W}$ is a closed set in $X$, so the set $p \left( \overline{W} \right)$ is a closed set in $Y$. Thus $p \left( \overline{W} \right)$ is a closed set in $Y$ containing $p(W)$. So we must have 
  $$ \overline{p(W)} \subset p \left( \overline{W} \right). \tag{2} $$
From (1) and (2) we obtain 
  $$ p \left( \overline{W} \right) = \overline{p(W)}. \tag{3} $$
As $p^{-1}(B) \subset W$ and as $p$ is surjective, so 
  $$ B = p\left( p^{-1}(B) \right) \subset p(W),   $$
  that is,
  $$ B \subset p(W). \tag{4} $$
As $\overline{W} \subset p^{-1}(U)$ by (A) above, and as $p$ is surjective, so 
  $$ p \left( \overline{W} \right) \subset p \left( p^{-1}(U) \right) = U, $$
  that is,
  $$ p \left( \overline{W} \right) \subset U. \tag{5} $$
Thus from (3), (4), and (5) we obtain
  $$ B \subset p(W) \subset p \left( \overline{W} \right) = \overline{p(W)} \subset U. \tag{6} $$
Now suppose that $y \in Y \setminus p(X \setminus W)$. Then $y \in Y$ and $y \not\in p(X \setminus W)$, and since $p \colon X \to Y$ is surjective, $y = p(x)$ for some point $x \in X$ such that $x \not\in X \setminus W$ and hence $x \in W$, which implies that $y \in p(W)$. Therefore 
  $$ Y \setminus p(X \setminus W) \subset p(W). \tag{7a}  $$

So what next? How to proceed from here? If $p$ were bijective or if $p$ were also an open map, then the proof would go through easily. But here $p$ is only surjective, and $p$ is closed.
How to use the hint given by Munkres?
PS:

Now as $p^{-1}(B) \subset W$ from (A) above and as $W \subset X$, so 
  $$ p^{-1}(B) \cap (X\setminus W) = \emptyset. $$ 
  So if $y \in B \cap p(X \setminus W)$, then $y = p(x)$ for some element $x \in X \setminus W$, and then $x \in p^{-1}(B) \cap (X \setminus W)$, which contradicts the fact that $p^{-1}(B)$ and $X \setminus W$ are disjoint. Therefore 
  $$ B \cap p(X \setminus W) = \emptyset, $$
  and this, together with the fact that $B \subset Y$, implies that 
  $$ B \subset Y \setminus p(X \setminus W). \tag{7b}$$
From (7a) and (7b) above we obtain
  $$ B \subset Y \setminus p(X \setminus W) \subset p(W). \tag{7} $$
Now as $W$ is an open set in $X$, so $X \setminus W$ is closed in $X$, and since $p \colon X \to Y$ is a closed map, the image set $p(X \setminus W)$ is a closed set in $Y$. Therefore the set $Y \setminus p(X \setminus W)$ is an open set in $Y$. Let us put 
  $$ V \colon= Y \setminus p(X \setminus W). \tag{Definition 0} $$
Then $V$ is an open set in $Y$, and  (7) becomes 
  $$ B \subset V \subset p(W), $$
  and this together with (6) yields
  $$ B \subset V \subset p(W) \subset \overline{p(W)} \subset U. \tag{8} $$
Now since $V \subset p(W)$, therefore we have 
  $$ \overline{V} \subset \overline{p(W)},$$ 
  and this together with (8) gives 
  $$ B \subset V \subset \overline{V} \subset U.$$
Thus we have shown that 
(1) one-point sets in $Y$ are closed, and 
(2) for any closed set $B$ in $Y$ and for any open set $U$ in $Y$ such that $B \subset U$, there exists an open set $V$ in $Y$ such that $B \subset V$ and $\overline{V} \subset U$. 
Hence by Lemma 31.1 (b) in Munkres $Y$ is normal. 

Is this proof sound enough now? If so, is my reasoning clear enough too? Or are there still any lacks and gaps left?
 A: The most straightforward way is just to take the bull by the horns and go direct:

That $Y$ is $T_1$ is, as you noticed, simply a consequence of closedness plus ontoness: if $p(x)=y$ then $\{y\}= p[\{x\}]$, the image of a closed set in $X$.
If $C$ and $D$ are closed and disjoint in $Y$, $C'=f^{-1}[C]$ and $D'=f^{-1}[D]$ are closed (by continuity) and disjoint in $X$ so have disjoint open neighbourhoods $C'\subseteq U'$ and $D' \subseteq V'$. Then by closedness of $f$, $U= Y\setminus f[X\setminus U']$ and $V= Y\setminus f[X\setminus V']$ are open in $Y$ and simple set theory shows that $U \cap V=\emptyset$ and $C \subseteq U$ and $D \subseteq V$, as required.

Cf. my answer here or my proof of Munkres' hint here for inspiration.
IMHO you make it way too complicated...
Added Some more explanation of the arguments involved:

*

*$U \cap V=\emptyset$: suppose not, then we have $y \in U \cap V$. As $f$ is surjective we write $y=f(x)$ for some $x \in X$. If $x \in X\setminus U'$  this would imply $f(x) \in f[X\setminus U']$ so $f(x) \notin U$, contradiction. So $x \notin X\setminus U'$ or $x \in U'$. Entirely similar is the argument that $x \in V'$, but then we contradict $U'\cap V'=\emptyset$. This final contradiction shows that $U$ and $V$ are indeed disjoint. We could also applied de Morgan and ontoness for a more "algebraic" proof:

$$
\begin{align}
U \cap V &= \left( Y\setminus f \left[X\setminus U'\right] \right) \cap \left( Y \setminus f \left[X \setminus  V' \right] \right) \\ 
&= Y \setminus \left( f \left[ X\setminus U' \right] \cup f \left[ X\setminus V' \right] \right) \\ 
&= Y \setminus f \left[ (X\setminus U') \cup (X\setminus V') \right] \\ 
&=  Y\setminus f \left[ X \setminus (U' \cap V') \right] \\
&=  Y \setminus f\big[ X\setminus \emptyset \big] \\
&= \emptyset.
\end{align}
$$

*

*$C \subseteq U$: Take $y \in C$ and assume $y \notin U$. This means that $y \in f[X\setminus U']$, so that $y=f(x)$ for some $x \in X\setminus U']$, or $x \notin U'$. But as $y=f(x) \in C$, $x \in C' = f^{-1}[C]$ and so $x$ would contradict the inclusion $C'\subseteq U'$. So $y \in U$ as required. $D \subseteq V$ is entirely similar again.

Hope this makes the total argument clearer.
