Show that the topology $T$ is $T_1$. Let $T$ be the topology on $R$ generated by subbasic open sets of the form [a,b] for all irrational number a,b.
(a) Determine the closure of the sets:
(1) (e,$\pi$); (2) ($\sqrt{2}$ , 5]; (3) {$\pi$,$\pi$/2,$\pi$/3,.....}
(b) Show that the topology is $T_1$.
For showing the topology is $T_1$, my argument is since these [a,b] are closed sets, if we take any two points in $T$ their neighbourhood will be disjoint so it will not contain any other point of $T$. Hence, it is $T_1$.
For finding closure, we need to find limit points, so I guess answers are as follows:
(1) [e,$\pi$]; (2) [$\sqrt{2}$ , 5]; (3) {$\pi$,$\pi$/2,$\pi$/3,.....}$\cup$ {0}
 A: SiD has given a brief but complete answer to the specific questions. However, this space is a rather important example in general topology, so I thought that you might find it useful to see another way of looking at it.
Let $\mathscr{S}=\big\{[a,b]:a,b\in\Bbb R\setminus\Bbb Q\text{ and }a\le b\big\}$, the subbase generating the topology $\mathscr{T}$. The base for $\mathscr{T}$ generated by $\mathscr{S}$ is the family of all intersections of finite subsets of $\mathscr{S}$. The intersection of a finite family of closed intervals with irrational endpoints is either empty or a closed interval with irrational endpoints. 

Specifically, if $\big\{[a_k,b_k]:k=1,\dots,n\big\}\subseteq\mathscr{S}$, let $a=\max\{a_k:k=1,\dots,n\}$ and $b=\min\{b_k:k=1,\dots,n\}$; then $$\bigcap_{k=1}^n[a_k,b_k]=\begin{cases}\varnothing,&\text{if }a>b\\ [a,b],&\text{otherwise}\;.\end{cases}$$

Thus, $\mathscr{S}$ is a base for $\mathscr{T}$. This means that a set $U\subseteq\Bbb R$ is open if and only if for each $x\in U$ there are irrationals $a_x$ and $b_x$ such that $x\in[a_x,b_x]\subseteq U$. More generally, $x\in\operatorname{int}U$ if and only if there are irrationals $a_x$ and $b_x$ such that $x\in[a_x,b_x]\subseteq U$, where $\operatorname{int}U$ is the interior of $U$.
Note that if $a$ is any irrational number, the set $\{a\}=[a,a]$ is open: each irrational is an isolated point of $\langle\Bbb R,T\rangle$. Now suppose that $q\in\Bbb Q$, and that $q\in U\subseteq\Bbb R$. If $q\in\operatorname{int}U$, there are irrationals $a$ and $b$ such that $q\in[a,b]\subseteq U$; clearly $a\ne q\ne b$, so $a<q<b$, and therefore $q\in(a,b)\subseteq U$. Conversely, suppose that there is some open interval $(u,v)$ in $\Bbb R$ such that $q\in(u,v)\subseteq U$. (Note that $u$ and $v$ need not be irrational.) Then $u<q<v$, so there are irrationals $a$ and $b$ such that $u<a<q<b<v$, and we have $q\in[a,b]\subseteq U$ and hence $q\in\operatorname{int}U$. In short, $q\in\operatorname{int}U$ if and only if there is an open interval $(u,v)$ such that $q\in(u,v)\subseteq U$. This means that $\{(u,v):u<q<v\}$ is a nbhd base at $q$ in $\langle\Bbb R\mathscr{T}\rangle$ as well as in the usual Euclidean topology on $\Bbb R$.
We can sum this up by saying that rational numbers have their usual nbhds in this new topology, while each irrational becomes an isolated point. If $\mathscr{E}$ is the Euclidean topology on $\Bbb R$, then $\mathscr{T}$ is the smallest topology containing both $\mathscr{E}$ and $\big\{\{x\}:x\in\Bbb R\setminus\Bbb Q\big\}$; equivalently, $\mathscr{E}\cup\big\{\{x\}:x\in\Bbb R\setminus\Bbb Q\big\}$ is another subbase for $\mathscr{T}$. This is actually how the space $\langle\Bbb R,\mathscr{T}\rangle$ is usually presented. 
As I intimated at the beginning, it’s a very well-known space: it’s called the Michael line, after the late topologist Ernest Michael. Some of its more interesting properties are proved in this post from Dan Ma’s Topology Blog.
A: Observe that any interval of the form $(x_1, x_2)$ for $x_i$ irrational is a closed set in T, because it is the complement of the open set $ (-\infty, x_1] \cup [x_2, +\infty)$. Hence the set in (1) is already closed, and also the set in (2) is because it is the intersection of the family of closed sets $(\sqrt{2}, x)$ where $x$ is irrational and greater than 5. Finally, note that every set in the given subbase that is also a neighbourhood of $0$ intersects the set in (3), and this implies that this point belongs to the closure of the set. Observe that if $\xi$ is a point different from $0$ and $\pi / n$ for every $n$, then there exist irrational numbers $x_1, x_2$ such that $\pi /n < x_1 < \xi < x_2 < \pi / (n-1) \leq \infty$ for some $n$; hence $\xi$ does not belong to the closure, and your answer is right...
The topology T is $T_1$ because for any two distinct points $x, y$ there is an irrational $\xi$ between them, and the sets $(-\infty, \xi]$ and $[\xi, +\infty)$ are open sets which contain one of the two points and not the other.
