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.
Best Answer
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.
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 $\tau$.
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.
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:
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:
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
This is absurd: $x$ cannot be contained in both rays, as it cannot be both less than $0$ and greater than $1$.