Let $X$ be a Hilbert space. I want to prove that for every countable subset $A \subset X$ there is an orthonormal set S with $\overline{\text{lin}\ S}=\overline{\text{lin}\ A}$. (Here is $\text{lin}\ Y$ the linear span of $\ Y$ and $\overline{W}$ is the closure of $\ W$).
I have alredy proved that for every countable subtset of linear independent vectors $\{ x_1,x_2,\dots \}$ there is a countable orthonormal set $\{ e_1,e_2,\dots \}$ such that $\text{lin}\{ x_1,\dots,x_n \}=\text{lin}\{ e_1,\dots, e_n \} \ $ for all $n\in\mathbb{N}$ (Gram-Schmidt process).
How can I go now from any countable linear independent subset to a countable subset of $X$? If that is possible, I think I could prove the claim with Gram-Schmidt if the countable set is also linear independent.
Thanks for your answers!
You can start with vector $x_1$. If it is linearly independent (non-zero), you take the next vector $x_2$. If $\{x_1,x_2\}$ is l.d., you just throw away $x_2$ and take $x_3$. Otherwise, you apply G-S to $\{x_1,x_2\}$. You can keep going this way ad infinitum.Since the discarded vectors did not contribute to the linear span, the obtained set of orthonormal vectors is countable and has the same linear span as the original sequence.