2.4:

Let g be a fixed element of G - S. (Such an element exists because S is assumed to be proper.) Let x be an element of G. If x is in G - S, then x is in < G - S >. If x is in S, then x = (g)(g^-1x), and both g and g^-1x, and both g and g^-1 are elements of G - S. Indeed, if g^-1x is not in G - S, then it must be in S, but this implies g is in S, since g = x( g^-1x)^-1 and S is a subgroup. Since x is a product of elements of G - S, x is in < G - S >. Thus < G - S > = G.