Do you mean separable for the topology of uniform convergence? If so, the answer is No.

Given a bounded sequence , define a bounded continuous function f by if n is an integer, and interpolating linearly between integer points. That shows that the map is surjective, and continuous, from the space CB(R) (continuous bounded functions on the line, with the uniform norm) to the space l^∞ of bounded sequences. But it is well known that l^∞ is not separable. Hence neither is CB(R).

The space of bounded continuous functions on abounded closed intervalis separable, because you can use Weierstrass's theorem to approximate functions uniformly by polynomials with rational coefficients.