 Solution

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.

The square classes are open in $K_v^{\times }$. Indeed the strong form of Hensel lemma applied to $X^2-u$ at $1$ shows that $u$ is a square whenever $v(u-1)>2v(2)$. Thus any sufficiently close approximation $a$ to $d_v$ has $a/d_v$ square. By weak approximation for number fields, choose $a\in K$ satisfying this at every finite $v\in S$. Then the local extensions $K_v(\sqrt{a})$ are all unramified of degree two. At least one such condition ensures $a$ is not a square in $K$. If $S$ contains no finite place, impose the same condition at one auxiliary finite place.

Set $L=K(\sqrt{a})$. At every specified finite prime there is one prime of $L$ above it, with ramification index one and residue degree two. Hence $\boxed{\mathfrak{p}{\mathcal{O}}_L\text{ is prime for every finite }\mathfrak{p}\in S}$. The dyadic case is covered by the unramified construction, rather than by a nonexistent nonsquare residue unit in characteristic two.