Are open intervals or union of open intervals the only two possible elements in any basis of usual topology on $\mathbb{R}$? Can $\mathbb{R}$ ususal topology has a basis which contains elements other than open interval?
 A: Certainly. E.g. the basis consisting of all open sets, or of all sets that are the union of two disjoint intervals, or of all sets that are the union of countably many disjoint open intervals.
A: A basis for the usual topology on $\mathbb{R}$ is a collection of open sets which (1) covers $\mathbb{R}$, and (2) is such that given any two elements $B_1,B_2$ in the basis and a point $x_0\in B_1\cap B_2$ we can find a set $B$ in the basis such that $x_0\in B\subseteq B_1\cap B_2$.
So the slightly more interesting question in the OP's comments was: can we find a basis which does not contain any open intervals at all? Answer: yes.
Take the collection $\{B\cup B': B,B'\text{ are disjoint nonempty open intervals}\}$. It is easy to see that it satisfies the conditions.
To make it slightly more interesting, we can find a countable basis of this type. Eg let $I_{n,r}=(r-3/n,r+1/n)\cup(r+2/n,r+3/n)$ and take the collection of all $I_{n,r}$ where $r$ is rational and $n$ is a positive integer. 
Note that $(r-1/n,r+1/n)\subset I_{n,r}$. The $I_{n,r}$ are obviously open sets and obviously cover $\mathbb{R}$. So if we take $x_0\in I_{n,r}\cap I_{n',r'}$ then some interval $(x-2/m,x+2/m)\subset I_{n,r}\cap I_{n',r'}$. We pick any rational $s$ in the interval $(x-1/m,x+1/m)$ and $I_{m,s}\subset (x-2/m,x+2/m)\subset I_{n,r}\cap I_{n',r'}$
