Let $X$ be a normed space and $F\subseteq X$. How to prove ${^\perp}(F^\perp)$ is the closed linear span of $F$? 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!
Clearly $A^{\perp}$ and $^{\perp}B$ are linear subspaces of the respective space. Also both of them are closed as preimage of a closed set. So it is only left to show that $F\subseteq {}^{\perp}(F^{\perp})$. $\renewcommand{\Re}{\operatorname{Re}} \newcommand{\cls}{\operatorname{cls}}$
take $f\in F$ then by Definition of $F^{\perp}$ we have $$\langle f,g \rangle = 0 \quad\text{for all}\quad g\in F^{\perp}.$$ Well and this justifies that $f$ is in ${}^{\perp}(F^{\perp})$.
From this we have $\cls F \subseteq {}^{\perp}(F^{\perp})$, where $\cls F = \overline{\operatorname{span} F}$.
For the other inclusion we need Hahn-Banach
Note that for $f\in X'$ we have that $f(\operatorname{cls} F)$ is a subspace of $\mathbb{C}$, which means it is either $\mathbb{C}$ or $\{0\}$. Using this and the theorem for $A = \{x\}$ where $x\notin \operatorname{cls}F$ and $B = \operatorname{cls}F$ we get an $f\in X'$ and $\gamma_1,\gamma_2 \in \mathbb{R}$ such that \begin{align*} \Re f (\{x\}) \leq \gamma_1 <\gamma_2 \leq \Re f(\cls F) \end{align*} since the right-hand-side is a subspace of $\mathbb{R}$ is must be $\{0\}$ and consequently $f(\cls F)=\{0\}$. In particular $f(F)= 0$. So this $f\in X'$ has the property $$ f\in F^\perp \quad \text{and}\quad \langle x,f\rangle \neq 0. $$ Hence $x\notin {}^{\perp}(F^{\perp})$