= Solution
If $F$ is empty, the zero <linear functional> satisfies the conclusion vacuously. Otherwise choose one $f_0\in F$, put $\delta=\epsilon/2$, and form the <convex set>
$$
E=F+B(0,\delta)-f_0.
$$
It contains $B(0,\delta)$ because $f_0\in F$. The point $x=-f_0$ is outside $E$: membership would imply $0=f+u$ for some $f\in F$ with $\|u\|<\delta$, contradicting the assumed absence of points of $F$ in $B(0,\epsilon)$.
Apply part (i). There is a <continuous linear functional> $T$ with $T(-f_0)\geq1$ and $T(f+u-f_0)\leq1$ for every $f\in F$, $\|u\|<\delta$. Thus $T\ne0$ and
$$
Tf+Tu\leq1+Tf_0\leq0.
$$
For fixed $f$, take the supremum of $Tu$ over this <open ball>; it is $\delta\|T\|$. Consequently $Tf\leq-\delta\|T\|$ for every $f\in F$. The normalization
$$
\boxed{S=-\frac{T}{\delta\|T\|}}
$$
is a <continuous linear functional> satisfying $Sf\geq1$ for every $f\in F$. No <compactness>, closedness or attainment of the supremum on the <open ball> is needed.
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