Q:

why use open interval rather than closed interval as open sets for real line topology

The set of all closed intervals of the real line also satisfies the open sets axioms.
1.The empty set and X itself are open.
2.Any union of open sets is open.
3.The intersection of any finite number of open sets is open.
Why the standard real line topology is defined on open intervals rather than using closed intervals as open sets?

A:

Another point of view...
For any set $\mathcal{S}$ of subsets of $\mathbb{R}$, there is a smallest topology $\mathcal{T}$ on $\mathbb{R}$ such that $\mathcal{S} \subseteq \mathcal{T}$. See here for details. So, in principle, you can use any set of subsets of $\mathbb{R}$ to generate a topology. However, the topology obtained will usually be different from the "standard topology" on $\mathbb{R}$, i.e. the one generated by the set of all open intervals.
In particular, let's see what topology $\mathcal{T}$ is generated by the set of all closed intervals of $\mathbb{R}$. For any $x \in \mathbb{R}$ we know $[x,x+1]$ and $[x-1,x]$ are closed intervals, so $[x,x+1],[x-1,x] \in \mathcal{T}$. However, $\mathcal{T}$ is a topology, so it is closed under finite intersections. Thus $\{x\} = [x,x+1] \cap [x-1,x] \in \mathcal{T}$. So, $\mathcal{T}$ contains every singleton $\{x\}$ where $x \in \mathbb{R}$. Now, using that $\mathcal{T}$ is closed under arbitrary unions, you should be able to see that every subset of $\mathbb{R}$ belongs to $\mathcal{T}$. That is, the topology $\mathcal{T}$ on $\mathbb{R}$ generated by the set of closed intervals is the entire power set of $\mathbb{R}$ -- also known, in this context, as the discrete topology on $\mathbb{R}$. 
So, in conclusion, using the closed intervals to generate a topology on $\mathbb{R}$ is not exactly "forbidden", but it does lead to a topology on $\mathbb{R}$ which is drastically different from (and much less useful than) the standard topology. 

