Can a Closed interval be a basis for Usual topology on$R$? I was thinking if $B$$=${$[a,b]$ , $a$ $\neq$ $b$ and a,b belong to $R$} is a basis for Usual topology on $R$.
I could not find a problem in this basis satisfying the two properties of basis.
Edit: if the answer to the above question is yes, then that fundamentally contradicts the fact that closed sets are not open in usual topology on $R$. How to  prove either of the above questions wrong?

The set $B$ will not form a basis for any topology on $\mathbb{R}$. Recall that a basis must have the property that for any $B_1, B_2 \in B$ and $x \in B_1 \cap B_2$, there exists $B_3 \in B$ such that $x \in B_3 \subset B_1 \cap B_2$. This is not satisfied by $B$, since $[a,b] \cap [b,c] = \{b\}$, which contains no element of $B$. 
If instead we take $B$ to be be a subbasis, then $B$ will generate the discrete topology on $\mathbb{R}$. Again, this follows from the fact that if $b \in \mathbb{R}$, then $[a,b] \cap [b,c] = \{b\}$ will be an element of the topology generated by $B$. 
