It is enough to prove that D(Rn) is dense in S′(Rn). Let λ be a continuous linear form on S′(Rn) which is zero on D(Rn). Since the space S(Rn) is reflexive, there exists a function φ∈S(Rn) such that λ(f)=⟨f,φ⟩ for every f∈S′(Rn). Hence
0=λ(ψ)=⟨ψ,φ⟩=∫Rnψφdμ
for every ψ∈D(Rn). The measure φ⋅μ on Rn is therefore zero (prop. 6 of IV, p. 206), whence φ=0. The proposition then follows from the Hahn–Banach theorem (EVT, II, p. 46, cor. 1).
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.