Prove that:
Let $(V, \Vert.\Vert)$ be a normed vectorspace. Then $V$ is separable iff $V$ has a total countable subset
Relevant definitions:
A total subset in a normed vector space $V$ is a subset $T \subseteq V$ such that $\overline{span T} = V$
$V$ is called separable iff it has a countable dense subset
My attempt:
$\boxed{\Rightarrow}$ Let $A$ be a countable dense subset in $V$. Then we have $\overline{A}= V$ and hence $span \overline{A} = V$. Now, because addition and scalar multiplication in vector spaces are continuous functions, and for a continuous function $f$ we have $f(\overline{A}) \subseteq \overline{f(A)}$, it follows that $\overline{span A} = V$ and hence $A$ is a countable total subset of $V$
$\boxed{\Leftarrow}$ Let $A$ be a total countable total subspace in $V$. Then $\overline{span A} = V$.
I'm stuck here. It is clear that we won't be able to prove that $span(A)$ is countable, because easy counterexamples can be found. So my idea was to show that $A$ is dense in $V$, given the denseness of $span(A)$ in $V$, but I was unsuccesful in doing this.
Is what I did correct? How can I complete the proof?
Hint: Recall how the separabiity of $C[0,1]$ follows from the Weierstrass approximation theorem.