= Solution
The <idele group> is the restricted product
$$
J_K=\prod_v' K_v^\times=\{(x_v):x_v\in K_v^\times,\ x_v\in\mathcal O_v^\times\text{ for almost every finite }v\},
$$
with componentwise multiplication and the <restricted product topology on the idele group> relative to the unit groups at finite places. We need <weak approximation for number fields> only at the finitely many specified coordinates, not density in the full restricted-product topology.
Here is a proof including the Archimedean places. At each finite place in $S$, choose $a_v\in K$ sufficiently close to $x_v$, using the definition of the completion. Choose a nonzero $d\in\mathcal O_K$ clearing their denominators. Pick powers $\mathfrak p_v^{m_v}$ so that $|z/d|_v<\varepsilon$ whenever $z\in\mathfrak p_v^{m_v}$. Put $I=\prod_{v\in S,\ v\text{ finite}}\mathfrak p_v^{m_v}$. For each positive rational integer $M$ coprime to the rational primes below these places, the <Chinese remainder theorem> gives $b_M\in\mathcal O_K$ with $b_M\equiv Md a_v\pmod{\mathfrak p_v^{m_v}}$. All of $b_M+I$ satisfy the same congruences.
Under the <Minkowski embedding of a number field>, $I$ is a full lattice in $\mathbb R^{r_1}\times\mathbb C^{r_2}$. A fundamental parallelepiped is bounded, so every real vector is within a fixed distance of this lattice, uniformly over its translates. Choose a target vector whose coordinates at the specified Archimedean places are those of $Md x_v$, assigning arbitrary values at the other Archimedean places. There is $b\in b_M+I$ within that fixed distance of the target. Dividing by $Md$ makes all specified Archimedean errors tend to zero as $M\to\infty$ through the allowed integers. At finite places $M$ is a unit, so the earlier congruence bounds persist. By taking the initial approximations sufficiently close, the ultrametric inequality gives the requested errors there too. \b[Thus some $\boxed{y=b/(Md)\in K}$ satisfies every required inequality.] If there are no finite places, set $I=\mathcal O_K$ and omit the congruences.
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