= Solution
Let
$$
m=\frac{v+w}{2},
\qquad
R(z)=v+w-z=2m-z.
$$
The map $R$ is reflection about $m$ and is an isometry. It interchanges $v,w$, so it preserves $S_1^{vw}$. Inductively, if it preserves $S_{n-1}^{vw}$, then it preserves its <diameter of a metric space>[diameter] and all distances in the defining condition for $S_n^{vw}$; hence it preserves every $S_n^{vw}$.
The midpoint belongs to $S_1^{vw}$ because
$$
\|m-v\|=\|m-w\|=\frac12\|v-w\|.
$$
Suppose $m\in S_{n-1}^{vw}$. For every $z\in S_{n-1}^{vw}$, reflection invariance gives $R(z)\in S_{n-1}^{vw}$ and
$$
\|m-z\|
=\frac12\|R(z)-z\|
\leq\frac12\operatorname{diam}(S_{n-1}^{vw}).
$$
Thus $m\in S_n^{vw}$, so all the sets are nonempty and contain $m$.
Put $D_n=\operatorname{diam}(S_n^{vw})$. If $x,y\in S_n^{vw}$ with $n\geq2$, then $y\in S_{n-1}^{vw}$ and the defining property of $x$ gives
$$
\|x-y\|\leq\frac12D_{n-1}.
$$
Therefore
$$
D_n\leq\frac12D_{n-1},
\qquad
D_n\leq2^{1-n}D_1\longrightarrow0.
$$
If $z$ lies in every $S_n^{vw}$, then both $z$ and $m$ lie in $S_n^{vw}$, so
$$
\|z-m\|\leq D_n\longrightarrow0.
$$
Hence $z=m$. We have proved the <metric extraction of a midpoint by shrinking diameters>:
$$
\boxed{\bigcap_{n\geq1}S_n^{vw}
=\left\{\frac{v+w}{2}\right\}.}
$$
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