Problem 17(v) Topology Without Tears (pg 96) 
The Question



My Understanding(final edit)
(My proof will probably suck,but it is the best I can do)
We are required to prove that $R\setminus C \in \tau$  and  $R\setminus A\in\tau$ are Hausdorff but not regular.
Assumptions
1.Let X be the set,of all real numbers and
$ S=\{1/n:n \in N\}$
2.Define a set $C\subseteq\Bbb R$ to be closed if $C=A \cup T$. is $A$ is closed in the Euclidean topology on $\Bbb R$. $T\subset S$
If C is closed by Def 1.2.4 R\C is open. Similarly if A is closed
in Euclidean space A\C is open
But T$\subset S $. Then every point of S is an isolated point, since a sufficiently small interval about 1/n doesn’t contain 1/m for any integer m$\ne$ n, and S has no interior points. The set of boundary points ofS, S ∪ {0}. The point 0 $\notin$ S is the only accumulation point ofS, since every open interval about 0 contains 1/n for sufficiently large n.
 Source
This implies that $\bigcup_{n\in N} S\cup {0}=[0,1]$
Thus R\S=$(-\infty,0) \cup (1,\infty$)
Next l will show it is a topology
$\emptyset$,X $\in\tau$, since R\C and R\A are open
and byProp 1.2.2 (ii) & (iii) are satisfied cause both sets
are open Thus R\C,R\A $\in\tau$ Finally R\S by prop 1.2.2 (ii)
is satisfied.
Thus C=A $\cup T \in \tau$
Then we can find x$\in (-\infty,0)\subset$ R\S
and y$\in. (1,\infty)\subset $R\S so that
-$\infty$<x<0 and 1<y<$\infty$
.
Let U=(-$\infty$,0) V=(1,$\infty$) Then U $\cap V=\emptyset$
So $\tau $ on R is Hausdorff and since O$\notin S$ we can’t
separate T$\in\tau$ so it is not regular.
 A: 
We are required to prove that $X\setminus C \in \tau$  and  $R\setminus A\in\tau$ are Hausdorff but not regular.

This is incorrect. What you are supposed to prove is that $\tau$ is a topology, and that the space $\langle\Bbb R,\tau\rangle$ is Hausdorff but not regular. You should begin by showing that $\tau$ really is a topology on $\Bbb R$, so I’ll skip down to that part of your answer.

Next l will show it is a topology
$\emptyset$,X $\in\tau$, since R\C and R\A are open

What are $C$ and $A$? What do they have to do with $\varnothing$ and $X$? You have asserted that $\varnothing$ and $X$ are in $\tau$, but what follows since has no identifiable connection with this assertion and therefore does not justify it.
Let $\mathscr{C}$ be the collection of sets that are defined in the problem to be closed: a set $C\subseteq\Bbb R$ is in $\mathscr{C}$ if and only if there are a set $A$ that is closed in the Euclidean topology and a set $T\subseteq S$ such that $C=A\cup T$. To show that $\Bbb R\in\tau$, you must show that it is the complement of some member of $\mathscr{C}$, i.e., that $\Bbb R\setminus\Bbb R\in\mathscr{C}$. $\Bbb R\setminus\Bbb R=\varnothing$; is $\varnothing\in\mathscr{C}$? Yes: $\varnothing$ is closed in the Euclidean topology, $\varnothing\subseteq S$, and $\varnothing=\varnothing\cup\varnothing$, so $\varnothing\in\mathscr{C}$, and therefore $\Bbb R=\Bbb R\setminus\varnothing\in\tau$. And $\varnothing=\Bbb R\setminus\Bbb R$, so to show that $\varnothing\in\tau$, you must show that $\Bbb R\in\mathscr{C}$. This is also true: $\Bbb R$ is closed in the Euclidean topology on $\Bbb R$, $\varnothing\subseteq S$, and $\Bbb R=\Bbb R\cup\varnothing$, so $\Bbb R$ meets the criteria to belong to $\mathscr{C}$, and $\varnothing=\Bbb R\setminus\Bbb R$ is therefore in $\tau$.

and byProp 1.2.2 (ii) & (iii) are satisfied cause both sets
are open Thus R\C,R\A $\in\tau$ Finally R\S by prop 1.2.2 (ii)
is satisfied.

This is nonsense, I’m afraid. Proposition $\mathbf{1.2.2}$ says that if you already have a topological space, then it satisfies these three conditions, but you don’t yet know that $\langle\Bbb R,\tau\rangle$ is a topological space: that’s what you’re trying to prove. To do so, you must show that $\tau$ satisfies the requirements of Definitions $\mathbf{1.1.1}$. The first two of these are identical to (i) and (ii) in Proposition $\mathbf{1.2.2}$, and the third requires just that the intersection of any two open sets be open. I proved above that $\tau$ satisfies (i), but (ii) and (iii) remain completely unjustified.

Thus C=A $\cup T \in \tau$

Since you never specified what $C,A$, and $T$ are, this is meaningless. We can guess, however, that you’re referring to the definition of closed sets given in the problem, in which case it is in general false: if $C\in\mathscr{C}$, then by definition $\Bbb R\setminus C$ is in $\tau$, not $C$, and whether or not $C$ is in $\tau$ is irrelevant to the task of showing that $\tau$ is a topology.
To prove that $\tau$ satisfies (ii), you must start with an arbitrary family $\mathscr{U}\subseteq\tau$ and show that $\bigcup\mathscr{U}\in\tau$. If $U\in\mathscr{U}$, then by definition $\Bbb R\setminus U\in\mathscr{C}$, so there are a Euclidean closed set $A_U$ and a $T_U\subseteq S$ such that $\Bbb R\setminus U=A_U\cup T_U$, and hence $U=\Bbb R\setminus(A_U\cup T_U)$. Then
$$\begin{align*}
\bigcup\mathscr{U}&=\bigcup_{U\in\mathscr{U}}\big(\Bbb R\setminus(A_U\cup T_U)\big)\\
&=\Bbb R\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\,,
\end{align*}$$
so to show that $\mathscr{C}\in\tau$, you must show that $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\in\mathscr{C}$. Let $A=\bigcap_{U\in\mathscr{U}}A_U$; this is an intersection of Euclidean closed sets, so it is Euclidean closed. Clearly $A\subseteq\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$; what other points might be in $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$? Any other point of $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$ must belong to $S$, so $A\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\subseteq S$. Thus, if we let $T=A\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$, we have $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)=A\cup T$, where $A$ is Euclidean closed, and $T\subseteq S$, and hence $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\in\mathscr{C}$, as desired. This shows that $\tau$ is closed under arbitrary unions, i.e., that it satisfies condition (ii).
Finally, you have to show that if $U,V\in\tau$, then $U\cap V\in\tau$. By definition there are Euclidean closed sets $A_U$ and $A_V$ and sets $T_U,T_V\subseteq S$ such that $U=\Bbb R\setminus(A_U\cup T_U)$ and $V=\Bbb R\setminus(A_V\cup T_V)$. Thus,
$$\begin{align*}
U\cap V&=\big(\Bbb R\setminus(A_U\cup T_U)\big)\cap\big(\Bbb R\setminus(A_V\cup T_V)\big)\\
&=\Bbb R\setminus\big((A_U\cup T_U)\cup(A_V\cup T_V)\big)\\
&=\Bbb R\setminus\big((A_U\cup A_V)\cup(T_U\cup T_V)\big)\,.
\end{align*}$$
$A_U\cup A_V$ is a Euclidean closed set, and $T_U\cup T_V\subseteq S$, so $$(A_U\cup A_V)\cup(T_U\cup T_V)\in\mathscr{C}\,,$$ and its complement $U\cap V$ is therefore in $\tau$. Thus, $\tau$ satisfies condition (iii) of Definitions $\mathbf{1.1.1}$ and is therefore a topology on $\Bbb R$.
The next step is to prove that $\langle\Bbb R,\tau\rangle$ is Hausdorff. For this you must let $x$ and $y$ be arbitrary distinct points of $\Bbb R$ and show that there are disjoint $U,V\in\tau$ such that $x\in U$ and $y\in V$. You can do this by considering cases — both points in $S$, exactly one of $x$ and $y$ in $S$, or both points in $\Bbb R\setminus S$ — but this is unnecessary. Any two distinct points of $\Bbb R$ have disjoint Euclidean open nbhds, and as Henno Brandsma proves in the second paragraph of his answer, every Euclidean open set is in $\tau$, so automatically any two distinct points of $\Bbb R$ have disjoint $\tau$-open nbhds.
The last step is to show that $\langle\Bbb R,\tau\rangle$ is not regular. For this you write:

since O$\notin S$ we can’t
separate T$\in\tau$ so it is not regular.

If by ‘O’ you mean $0$, you may possibly have identified the counterexample to regularity, but what you’ve written doesn’t make any sense. Here is an actual argument.
First, by definition $S=\varnothing\cup S$ is $\tau$-closed, since $\varnothing$ is Euclidean closed and $S\subseteq S$. And $0\notin S$, so if $\langle\Bbb R,\tau\rangle$ were regular, there would be disjoint $U,V\in\tau$ such that $0\in U$ and $S\subseteq V$. Thus, you can prove that $\langle\Bbb R,\tau\rangle$ is not regular by proving that no such $U$ and $V$ exist.
Suppose, then, that $U,V\in\tau$, $0\in U$, and $S\subseteq V$; you need to prove that $U$ and $V$ cannot be disjoint. By definition there are Euclidean closed sets $A_U$ and $A_V$ and sets $T_U,T_V\subseteq S$ such that
$$U=\Bbb R\setminus(A_U\cup T_U)=(\Bbb R\setminus A_U)\cap(\Bbb R\setminus T_U)\,,$$
and similarly,
$$V=\Bbb R\setminus(A_V\cup T_V)=(\Bbb R\setminus A_V)\cap(\Bbb R\setminus T_V)\,.$$
$\Bbb R\setminus A_U$ is a Euclidean open set; call it $W_U$. Then
$$U=W_U\cap(\Bbb R\setminus T_U)=W_U\setminus T_U\,.$$
Similarly $\Bbb R\setminus A_V$ is Euclidean open and if we call it $W_V$, we have $V=W_V\setminus T_V$. But $V\supseteq S$, so $T_V=\varnothing$, and $V=W_V$. In other words, $V$ is already a Euclidean open nbhd of $S$.
$0\in U\setminus S\subseteq U\setminus T_U$, so $0\in W_U$. $W_U$ is a Euclidean open nbhd of $0$, so there are $a,b\in\Bbb R$ such that $0\in(a,b)\subseteq W_U$. Note that $(a,b)\setminus S\subseteq W_U\setminus T_U=U$.
Clearly $0<b$, so there is a positive integer $n$ such that $\frac1n<b$. $\frac1n\in S$, and $V$ is a Euclidean open nbhd of $S$, so there are $c,d\in\Bbb R$ such that $\frac1n\in(c,d)\subseteq V$. Let $r=\max\{\frac1{n+1},c\}$, and let $x=\frac12\left(r,\frac1n\right)$; then $\frac1{n+1}<x<\frac1n$, so $x\notin S$, and therefore $x\in(a,b)\setminus S\subseteq U$. Thus, $x\in U\cap V$, and $U$ and $V$ are therefore not disjoint. Since $U$ and $V$ were arbitrary $\tau$-open nbhds of $0$ and $S$, we’ve shown that $0$ and the $\tau$-closed set $S$ cannot be separated by disjoint open sets and hence that $\langle\Bbb R,\tau\rangle$ is not regular.
That completes the proof, but I want to comment a few other things that you wrote. Early on you wrote this:

But T$\subset S $. Then every point of A is an isolated point, since a sufficiently small interval about 1/n doesn’t contain 1/m for any integer m$\ne$ n, and A has no interior points. The set of boundary points of A is A ∪ {0}. The point 0 $\notin$ A is the only accumulation point of A, since every open interval about 0 contains 1/n for sufficiently large n.
 Source
This implies that $\bigcup_{n\in N} S\cup \{0\}=[0,1]$
Thus R\S=$(-\infty,0) \cup (1,\infty$)

It implies nothing of the kind. $\bigcup_{n\in\Bbb N}S$ is simply $S$, and $S\cup\{0\}$ certainly is not the entire closed unit interval. Then you say that we can find

x$\in. (1,\infty)\subset $R\S so that
-$\infty$<x<0 and 1<x<$\infty$
Clearly x is contained in both intervals

This is absurd: $x$ cannot be contained in both rays, as it cannot be both less than $0$ and greater than $1$.
A: You should perhaps start by proving that this definition of closed sets of $\mathcal{T}$ as sets of the form $A \cup S$, where $A$ is Euclidean closed and $S \subseteq T = \{\frac1n\mid n \in \Bbb N\}$ is a valid one, in that it obeys the axioms for closed sets for a topology.
Secondly note that if $O$ is Euclidean open, $\Bbb R\setminus O$ is Euclidean closed, so also closed in $\mathcal{T}$ (take $S=\emptyset$ in the definition) and hence $O$ is open in $\mathcal{T}$. So because we can separate any two points of $\Bbb R$ in the Euclidean topology, we can use these same open sets to separate them in $(\Bbb R, \mathcal T)$. So the newly defined space is still Hausdorff.
Moreover, $T$ is closed in $\mathcal T$ (take $A=\emptyset, S=T$ in the definition) and we can show that $0 \notin T$ cannot be separated from $T$ in $\mathcal T$ and this will show that $(\Bbb R, \mathcal T)$ is not regular.
Try to focus on this giving last proof, and forget about the (incomprehensible, at least for me) considerations in your post. The proof could go by contradiction, and should at least use that $0 \in \overline{T}$ in the Euclidean topology...
