Attempted Solution: Let $S = {(a,∞) : a ∈ R}∪{(−∞,b) : b ∈ R}$, the set of all infinite rays in R. Then S is clearly not a basis for the standard topology on R. For example: (0,1) is open in R, but I can’t find an infinite ray containing 1/2 which is contained in (0,1). But S is a sub-basis, for $(a,b) = (a,∞) ∩ (−∞,b)$, so if U is an open set containing the point x, and $x ∈(a,b) ⊂ U$, we have $x ∈ (−∞,b)∩(a,∞) ⊂ U$.
Let $L=\{\cap_{i=1}^n S_i:S_i\in S,n\in N\}$
Let $A\in L$ so $A$ is the intersection of two sets of $S$. As $S$ is the union of LHS and RHS, $A$ can be $A=(-\infty,c) \, c\leq b$ or $A=(d, \infty) \, d\geq a$ or $A=(x,y) \, x<y$ or $A=\emptyset$, since $L$ contains every open interval then is a basis for the usual topology on $R$.