Let $A=\{\frac{1}{n}:n\in\mathbb{N}\}$ and $\tau=\{G\setminus B:G \ \text{is an open subset of $\mathbb{R}$ and}\ B\subseteq A \}$. I have already proved that $\tau $ is a topology on $\mathbb{R}$. Now I need to prove that $(\mathbb{R},\tau)$ is Hausdorff but not regular. Obviously $\tau_{Euc} \subseteq \tau$, where $\tau_{Euc}$ is the Euclidean topology on $\mathbb{R}$. Moreover $(\mathbb{R},\tau_{Euc})$ is Hausdorff, whence $(\mathbb{R},\tau)$ is Hausdorff. The following is my proof of the latter part.
Suppose $(\mathbb{R},\tau)$ is regular. $A=\mathbb{R}\setminus(\mathbb{R}\setminus A)$; therefore $A$ is $\tau$-closed. But $0\notin A$. So by regularity, there exists a neigborhood $U$ of $0$ and $V\in\tau$ such that $A\subseteq V$ and $U\cap V=\emptyset$. Say $U=G\setminus B$, where $G\in\tau_{Euc}$ and $B\subseteq A$, and $V=H\setminus C$, where $H\in\tau_{Euc}$ and $C\subseteq A$. But since $A\subseteq V=H\setminus C$ we have $C=\emptyset$. Therefore $V=H\in\tau_{Euc}$. But $H\cap G\neq\emptyset$ (as $0\in G$ and therefore $(-\epsilon,\epsilon)\subseteq G $ for some $\epsilon>0,$ and there exists $\frac{1}{n_{\epsilon}}\in\mathbb{N}$ such that $0<\frac{1}{n_{\epsilon}}<\epsilon$ and $\frac{1}{n_{\epsilon}}\in H$). Also $H\cap(G\setminus B)=\emptyset$. Hence $G\subseteq B$ (as $H\cap G\neq\emptyset$ ); contradiction. Hence $(\mathbb{R},\tau)$ is not regular.
Is this proof alright? Please help. Thank you.
Once you have the $\frac{1}{n_\varepsilon} \in U$ you are done : $\frac{1}{n_\varepsilon} \in A \subseteq V$, so right away: $ U \cap V \neq \emptyset$ contradiction. The rest is fine, except (nitpick) I wouldn't introduce that many distinct letters (like $G,H,C$ in the notation. Maybe write $U = U' \setminus A_U$, where $U'$ is usual open and $A_U \subseteq A$, etc.
But the point is that $\frac{1}{n_\varepsilon}$ need not be in $U$, as we subtract members of $A$ from it.
We need to use openness of $V$ as well:
So suppose $0\in U = U' \setminus A_U, A\subseteq V$ with $U \cap V=\emptyset$, where $U'$ and $V$ are Euclidean open and $A_U\subseteq A$. We can even take $ A_U= A$ WLOG as it only makes $U$ smaller.
($V$ must be Euclidean open as we cannot miss points of $A$. ) then for some $n$ we have $\frac{1}{n} \in U'$ from being inside some $(-\varepsilon,\varepsilon)\subseteq U'$. Then $\frac{1}{n}\in V$ so there is some $\delta>0$ such that $(\frac{1}{n}-\delta,\frac{1}{n}+\delta) \subseteq V$ which means there is some point $p\notin A$ which lies in $(\frac{1}{n}-\delta, \frac{1}{n})$ and this $p$ then also lies in $(0,\varepsilon) \subseteq U'$. So $p \in U\cap V$ contradiction.