Complete set of vectors? The following excerpt is from the book Operators, Functions, and Systems, Vol. 1 by Nikolai K. Nikolski, pages 216-217


What is a "complete set of vectors" in the proof of Lemma 2.4.4?
Im putting my money on 
"A subset of the basis which is linearly independent and whose span is dense is called a complete set"
 A: If I understand correctly from this link, it seems that a set of basis vectors is said complete if and only if its elements span the all space. 
E.g. in $\Bbb R^{3}$ seen as a vector space of $\Bbb R$, we have the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$. The family $\{(1,0,0),(0,1,0)\}$ of basis vectors is not complete as it does not span all of $\Bbb R^{3}$.
In finite dimensional settings, the terms "basis" and "complete set" are equivalent. It may have a subtlety when the set is infinite dimensional. Anyway, for a book about functional analysis, it seems to be a confusing terminology, as completeness (in the topological sense) is an important property in many functional spaces.
For what you are "putting your money on", I guess it would make sense but asking that they are linearly independent is useless as, if they are a subset of the basis, they must be linearly independent.
You might be interested in this question from the physics community addressing the different notions of "completeness" for a family of vectors.
