 Solution

The idele group is the restricted productwith 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 $|z/d{|}_v<\epsilon$ 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 $b_M\equiv Md{a}_v(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$. 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. 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.