Let v∈Ω(E). One has ∥v∗∥=∥v∥⩽1 therefore the endomorphism 1E−vv∗ is positive. It is injective: if x∈ Ker(1E−vv∗), one has vv∗(x)=x, whence v∗(v(v∗(x)))=v∗(x) then v∗(x)=0, since 1E−v∗v is injective, and finally x=v(v∗(x))=0. The lemma follows.
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.