Let B be a bounded subset of D(Rn). The topology induced on B by the topology of S(Rn) coincides with the topology induced by D(Rn).
Since the inclusion of D(Rn) in S(Rn) is continuous, the topology on B induced by D(Rn) is finer than that induced by S(Rn). On the other hand, there exists a compact subset K of Rn such that B⊂CK∞(Rn). For every α∈Nn, one then has pα,K(φ)⩽qα,0(φ), which implies that the topology induced by S(Rn) is finer than that induced by D(Rn).
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.