Let H be such an intersection. The automorphism x*G/x maps chains of subgroups to chains of subgroups. In other words, conjugation preserves the lattice of subgroups of G, and maps the intersection of maximal subgroups onto itself, hence H is normal in G.
Let K be another proper subgroup of G. What is K join H? Push K up to a maximal subgroup of G, and H join K can only increase. Yet H is contained in every maximal subgroup. Therefore any subgroup of H, joined with any proper subgroup of G, yields a proper subgroup of G.
Let P be a sylow subgroup of H. Remember that conjugation by x performs an automorphism on H. This moves P to a sylow subgroup of H that is isomorphic to P. Since all sylow subgroups are conjugate, some y in H carries the image of P back to P. Write this as yxP/x/y = P. In other words, yx is in the normalizer of P.
Let K be the normalizer of P, and note that H join K includes y inverse times yx. Since x was arbitrary, H join K is all of G. Yet this is a contradiction, unless K = G. Therefore the normalizer of P is G, and P is normal in G.
Restrict to H, and P is normal in H. This holds for all sylow subgroups of H, and by the previous theorem, H is nilpotent.