Proving for each seperatble hilbert space exist complete sequence 
Let $H$ be a separable Hilbert Space. Prove that exists orthonormal complete sequence and give example for  one non-orthonormal sequence.

I thought taking orthonormal basis for $H$  denoted by $\{e_1,\dots ,e_n\}$ and then proving it's also complete, i.e $$(e_n,f)=0\Rightarrow f\equiv 0$$ $f$ can be represented as $f=\sum_{k=1}^\infty(e_k,f)e_k$ so I need to prove that $\forall k\in\mathbb{N}, (e_k,f)=0$ which as far as I see is given above, isn't it? Where am I using the separability?
 A: Suppose $H$ has a countably dense subset $\{ s_{n}\}_{n=1}^{\infty}$ of non-zero vectors. Using induction we may discard each $s_{n}$ which is a linear combination of the previous vectors in the sequence in order to arrive at a new sequence $\{ t_{n}\}_{n=1}^{\infty}$ of vectors for which the $\{ t_{1},t_{2},\cdots,t_{k}\}$ is linearly independent for each fixed $k=1,2,3,\cdots$. Then we may perform Gram-Schmidt on the resulting set to obtain an orthonormal sequence $\{ e_{1},e_{2},e_{3},\cdots\}$ such that the first $n$ elements of $\{ t_{k}\}$ spans the same subspace as the first $n$ elements of $\{ e_{k}\}$.
Automatically, finite linear combinations of the $e_{k}$ form a dense linear subspace of $H$ because this subspace includes the original dense subset $\{ s_{n}\}$, which is one equivalent to the orthonormal subset $\{ e_{k}\}_{k=1}^{\infty}$ being complete. A simple example of a non-orthonormal sequence is $\{ e_{1},e_{1}+e_{2},e_{1}+e_{3},e_{1}+e_{4},\cdots\}$. The inner product of any two of the elements of this sequence is $1$.
