Suppose $A$ is an uncountable complete set in a separable Hilbert space. Is it possible to find a complete countable subset $B$ of $A$?
As an example, take the uncountable set $A = \{ | \alpha \rangle \}$ $$ | \alpha \rangle = \sum_{n=0}^{\infty} \frac{\alpha^n}{n!} |n\rangle $$ where $\alpha$ is an arbitrary complex number. Every converging series $\{ \alpha_j\}$ of in the complex plane induces a complete countable set $\{| \alpha_j \rangle\}$. Is it always possible to reduce an uncountable complete set to a countable set that is complete?
My idea to show that a complete countable subset $B$ always exists looks like this (forgive me my physics notation):
Since the Hilbert space is separable, there exists an ONB denoted by $\{|n \rangle \}$. We denote the elements of $A$ by $| \alpha \rangle$. Since $A$ is complete, each basis element can be written like $$ | n \rangle = \sum_{i} a_{n,i} |\alpha_{n,i} \rangle $$ An element of the Hilbert space can be written as $$ | \psi \rangle = \sum_{n} c_n | n \rangle = \sum_n \sum_i c_n a_{n,i} | \alpha_{n,i} \rangle $$ The set $B = \{ |\alpha_{n,i} \rangle \}$ is complete and countable. Is there anything wrong with this line of thought?