= Solution
Choose a maximal subset $U\subset S^{n-1}$ whose distinct points have <Euclidean distance> greater than one. The <volumetric bound for Euclidean metric nets> makes the construction finite: the open <Euclidean balls> of radius $1/2$ about the points of $U$ are disjoint and all lie in $(3/2)B^n$, so
$$
|U|\,2^{-n}v_n\leq(3/2)^nv_n,\qquad |U|\leq3^n.
$$
Maximality means that $U$ is a <metric net> of radius one on the <unit sphere>. Thus every $\theta\in S^{n-1}$ has some $u\in U$ with $\lVert u-\theta\rVert\leq1$. Since both are <unit vectors>,
$$
u\cdot\theta=1-\frac12\lVert u-\theta\rVert^2\geq\frac12.
$$
Now intersect the corresponding <closed half-spaces>:
$$
\boxed{P=\bigcap_{u\in U}\{x:u\cdot x\leq1/2\}.}
$$
The <Cauchy-Schwarz inequality> shows that $(1/2)B^n\subset P$. Conversely, write any nonzero $x\in P$ as $x=t\theta$ and choose $u$ as above. Then $t/2\leq t(u\cdot\theta)=u\cdot x\leq1/2$, so $t\leq1$. Hence $P\subset B^n$, which also proves boundedness. This <convex polytope> has at most $|U|\leq3^n$ <facets>, since redundant inequalities can only reduce their number. \b[It meets the required inclusions with $r=2$.]
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