Prob. 4, Sec. 31, in Munkres' TOPOLOGY, 2nd ed: Satisfaction or otherwise of separation axioms as topology on a set become coarser or finer Here is Prob. 4, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ and $X^\prime$ denote a single set under two topologies $\mathscr{T}$ and $\mathscr{T}^\prime$, respectively; assume that $\mathscr{T}^\prime \supset \mathscr{T}$. If one of the spaces is Hausdorff (or regular, or normal), what does that imply about the other?

My Attempt:

We are given that 
  $$
X = \big( S, \mathscr{T} \big), \qquad \mbox{ and } \qquad  X^\prime = \left( S, \mathscr{T}^\prime \right), \tag{Definitions 0}
$$
  where $S$ is a non-empty set, and also that 
  $$
\mathscr{T}^\prime \supset \mathscr{T}. \tag{0}
$$
Hausdorffness: 
If $X$ is a Hausdorff space, then so is $X^\prime$. 
Let $x$ and $y$ be any two distinct points of $X^\prime$. As $x$ and $y$ are also distinct points of the Hausdorff space $X$, so there exist disjoint sets $U$ and $V$ of $X$ such that $x \in U$ and $y \in V$, and as $U, V \in \mathscr{T}$ and $\mathscr{T} \subset \mathscr{T}^\prime$, so we can conclude that $U, V$ are in $\mathscr{T}^\prime$ also. Thus $X^\prime$ is also a Hausdorff space.
However, if $X^\prime$ is a Hausdorff space, then $X$ need not be Hausdorff. 
For example, let 
  $$ S \colon= \{ a, b \}, $$
  and let 
  $$ 
\mathscr{T} \colon= \big\{ \emptyset, X \big\}, \qquad \mathscr{T}^\prime \colon= \mathscr{P}(X). 
$$
  Then $X^\prime = \left( S, \mathscr{T}^\prime \right)$ is of course a Hausdorff space, because the sets $\{ a \}$ and $\{ b \}$ are the disjoint open sets of $X^\prime$ containing the distinct points $a$ and $b$. However, $X = \big( S, \mathscr{T} \big)$ is not a Hausdorff space.
Regularity:
If $X = \big( S, \mathscr{T} \big)$ is a regular space, then $X^\prime = \left( S, \mathscr{T}^\prime \right)$ need not be regular. 
For example, let $S \colon= \mathbb{R}$, and let $\mathscr{T}$ denote the standard topology on $\mathbb{R}$ whereas $\mathscr{T}^\prime$ denote the $K$-topology on $\mathbb{R}$. 
To show that $\mathbb{R}$ with the standard topology is regular, we note that one-point sets in $\mathbb{R}$ are closed. Let $x \in \mathbb{R}$ and $U$ be any open set in $\mathbb{R}$ containing $x$. Then there exists an open interval $(a, b)$, where $a$ and $b$ are real numbers such that $a < b$, satisfying
  $$ x \in (a, b) \subset U. $$
  Now as 
  $$
a < x < b,
$$
  so we can find real numbers $c$ and $d$ such that 
  $$
a < c < x < d < b.
$$
  Then let us put
  $$
V \colon= (c, d).
$$
  This set $V$ is open in $\big( \mathbb{R}, \mathscr{T} \big)$, and also 
  $$
\overline{V} = [c, d] \subset (a, b) \subset U,
$$
  and hence 
  $$
\overline{V} \subset U.
$$
  Hence $\mathbb{R}$ with the standard topology is a regular space, by Lemma 31.1 (a) in Munkres.

Here is a Math Stack Exchange post of mine that is relevant.

However, $\mathbb{R}_K$ is not a regular space. Please refer to Example 1, Sec. 31, in Munkres.
On the other hand, if we let $\mathscr{T}^\prime$ denote the lower-limit topology on $\mathbb{R}$, then $\mathbb{R}_l \colon= \left( \mathbb{R}, \mathscr{T}^\prime \right)$ is a regular space. Please refer to Example 2, Sec. 31, in Munkres.
And, if $X^\prime = \left( S, \mathscr{S}^\prime \right)$ is regular, then $X = \big( S, \mathscr{T} \big)$ need not be regular.
For example, let $S \colon= \mathbb{R}$, let $\mathscr{T}^\prime \colon= \mathscr{P}(\mathbb{R} )$, and let $\mathscr{T}$ denote the $K$-topology on $\mathbb{R}$. Then $X^\prime$ is regular, but $X$ is not. Please refer to Example 1, Sec. 31, in Munkres.
On the other hand, if we let $\mathscr{T}$ be the standard topology on $\mathbb{R}$, then $X = \big( \mathbb{R}, \mathscr{T} \big)$ is a regular space.
Normality:
If $X = \big( S, \mathscr{T} \big)$ is normal, then $X^\prime = \left( S, \mathscr{T}^\prime \right)$ need not be normal.
For example, let $X$ denote the set $\mathbb{R}$ of real numbers with the standard topology, and let $X^\prime$ denote $\mathbb{R}$ with the $K$-topology. Then $X$, being metrizable, is normal by Theorem 32.2 in Munkres, but $X^\prime$ is not regular (by Example 1, Sec. 31, in Munkres) and therefore not normal.
And, if $X^\prime = \left( S, \mathscr{T}^\prime \right)$ is normal, then $X = \big( S, \mathscr{T} \big)$ need not be normal. 
For example, let $S \colon= \{ a, b, c\}$, and let 
  $$ \mathscr{T} \colon= \big\{ \emptyset, \{ a, b \}, \{ b, c \}, \{ b \}, X \big\} $$
  and 
  $$ \mathscr{T}^\prime \colon= \mathscr{P}(X). $$
Then $X^\prime$ is normal, because for any two disjoint closed sets $A$ and $B$ in $X^\prime$, we can take $A$ and $B$ themselves to be the disjoint open sets containing $A$ and $B$, respectively.
However, $X$ is not normal. the disjoint closed sets $\{ c \}$ and $\{ a \}$ have no disjoint open sets containing them.
Thus we can conclude that if $S$ is any non-empty set, and if $\mathscr{T}$ and $\mathscr{T}^\prime$ are any topologies on $S$ such that $\mathscr{T}^\prime \supset \mathscr{T}$, then the Hausdorffness, regularity, or normality of $X^\prime = \left( S, \mathscr{T}^\prime \right)$ implies nothing about the Huasdorffness, regularity, and normality, respectively, of $X = \left( S, \mathscr{T} \right)$. And, the regularity or normality of $X$ implies nothing about the regularity and normality, respectively, of $X^\prime$. However, the Hausdorffness of $X$ does imply the Hausdorffness of $X^\prime$. 

Is my solution correct in each and every detail? Or, are there issues? 
 A: It all looks fine, maybe you could rely more on earlier results, and reprove less, as a general tip to reduce the size of your write-ups.
Of course you can never conclude anything for the smaller topology from the property of the larger one here: any discrete space $(X,\mathcal{T}')$ is Hausdorff, regular, normal but we can take any suitable (automatically coarser) $\mathcal{T}$ on $S$ to show it need not be Hausdorff, normal or regular. No need for invention of new examples.
A finer topology than a Hausdorff can is always Hausdorff (because it's a "point-property"), but regular and normal can be lost because then we also introduce more closed sets that are to be checked for regularity and normality. The $K$-topology is standard for regularity (and thus also normality a fortiori), but for normality you could also have used the Sorgenfrey plane $\mathbb{R}_l \times \mathbb{R}_l$ as Munkres describes it) vs the standard plane. That fits nicely into Munkres' collection of favourite examples.
