In this proof line -3: "${^\perp}(F^\perp)$ is closed linear span of $F$ and this subspace of $X$ is separable." I know ${^\perp}(F^\perp)=\overline{\operatorname{span}(F)}$, but we only know $F$ is countable. How to show $span(F)$ is countable to satisfy ${^\perp}(F^\perp)$ is separable?
Here is some definitions: If $A\subseteq X$, then we denote $$A^{\perp}=\{x^*\in X^*:\langle a,x^*\rangle=0 \textrm{ for all }a\in A\}.$$ Similarly, if $B\subseteq X^*$, then we denote $$^{\perp}B=\{x\in X:\langle x,b^*\rangle=0 \textrm{ for all }b^*\in B\}.$$
Thank you!
The linear span of a countable set $F$ is separable : pick a countable dense set $D \subseteq \mathbb{F}$, say $\mathbb{Q}$ or $\mathbb{Q} + i\mathbb{Q}$ depending on whether the field of reals or complex numbers is used, and then the set
$$\{ \sum_{i=1}^n q_i \cdot x_i: n \in \mathbb{N} ,\forall_{1\le i \le n}: q_i \in D, x_i \in F\}$$
is countable (a union of images of the countable sets $\mathbb{Q}^n \times F^n$ over all finite $n$) and dense in $\operatorname{span}(F)$. And if $\operatorname{span}(F)$ is separable, so is its closure, obviously.