= 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. \b[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.
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