= Solution
The <complex Hahn-Banach theorem> extends every bounded complex-<linear functional> on a subspace to the whole normed space with exactly the same <norm>. Let $d=\operatorname{dist}(x_0,M)>0$ and define on $M+\mathbb Cx_0$
$$
g_0(y+\lambda x_0)=\lambda d.
$$
The representation is unique. The distance inequality $\|y+\lambda x_0\|\geq|\lambda|d$ gives $\|g_0\|\leq1$. Taking $y=-m$ with $\|x_0-m\|\downarrow d$ proves the reverse <norm> bound. Extend by the <Hahn-Banach theorem> to $g\in X'$; then \b[$g|_M=0$, $\|g\|=1$, and $g(x_0)=d\ne0$]. The labels (a)-(c) specify these simultaneous properties of this one functional.
Let $Y=\overline{\operatorname{span}F}$. Every functional vanishing on $F$ vanishes on $Y$ by linearity and continuity, so $Y\subseteq(F^\circ)_\circ$. If $x\notin Y$, the separation just proved gives a bounded functional zero on $Y$ and nonzero at $x$. It belongs to $F^\circ$ and excludes $x$ from its preannihilator. Therefore
$$
\boxed{(F^\circ)_\circ=\overline{\operatorname{span}F}.}
$$
For the dual-side counterexample take $X=\ell^1$, $X'=\ell^\infty$, and $G=c_0$. It is <norm> closed, since a uniform limit of sequences tending to zero still tends to zero. Every coordinate <unit vector> belongs to $G$, so the only $x\in\ell^1$ annihilated by every element of $G$ has all coordinates zero. Thus $G_\circ=\{0\}$ and
$$
\boxed{(G_\circ)^\circ=\ell^\infty\ne c_0=G.}
$$
The dual-side double annihilator corresponds to weak-star <closure>, which need not equal <norm> <closure>.
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