TITLE: Normed separable space, linearly independent $X_0 \subset X, \ \overline{linX_0} = X $
QUESTION [5 upvotes]: Could you tell me how to prove that 
a normed  space $X$ is separable $\iff$ there exists an at most countable set of linearly independent vectors $X_0 \subset X$ such that $ \ \text{lin} X_0$ is dense in$ X $?
I know how to prove that if a normed space is separable, then there exists such a set.
But I don't know what to do about the opposite direction. Could you help?
Thank you

REPLY [4 votes]: Given a countable set with dense span:
$$\#A_0\leq\mathfrak{n}:\quad\overline{\langle A_0\rangle}=X$$
Consider the rational span:
$$S_0:=\langle A_0\rangle_{\mathbb{Q}+\imath\mathbb{Q}}:\quad\#S_0\leq\mathfrak{n}$$
Its closure captures the linear span as:
$$a,a'\in A_0:\quad\|\lambda a+\lambda'a'-\lambda_na-\lambda'_na'\|\leq|\lambda-\lambda_n|\cdot\|a\|+|\lambda'-\lambda'_n|\cdot\|a'\|\to0$$
So it was already dense:
$$\langle A_0\rangle\subseteq\overline{\langle S_0\rangle}\implies X=\overline{\langle A_0\rangle}=\overline{\langle S_0\rangle}$$
(The trick here is that the closure is closed; cf. Kuratowski.)