The proof takes a dense set (array) in the space and takes out all the vectors who are a linear combination of its predecessors. It is claimed that with this procedure we get a countable array of linearly independent vectors and by using Gram-Schmidt orthogonalization we get the needed set.
I'm confused why must the resulting array be countable? What if there are only finitely many vectors who are linearly independent? I can't see the contradiction here.
The comment by @J.Loreaux actually answered my question (in the lemma from my lectures it was left out that the space is infinite dimensional):
"...If the space is infinite dimensional, the resulting set cannot be finite. This is because the original countable dense set must have infinitely many linearly independent vectors. Indeed, if it didn't, it would be contained in a finite dimensional subspace and then could not be a dense subset. The last claim follows because there is a hyperplane separating your finite dimensional subspace from any point not in the subspace (here I am assuming you are in a locally convex topological vector space, i.e. the topology is generated by a family of seminorms; this may be overkill)"