 Solution

Normalize each finite-place valuation to have image $\mathbb{Z}$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $a{b}^n$ adds $nv(b)$. A suitable finite set isOne can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero valuations, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.

Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a={\pi }^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the residue field, $X^n-\bar{u}$ is separable: its derivative $n{X}^{n-1}$ has no common zero with it.

Take a finite extension of the finite residue field containing all roots of all these finitely many reduced polynomials. The corresponding finite unramified extension of $K_v$ exists by the classification proved in Question 2. Hensel's lemma lifts each simple residue root to a root of the original polynomial in that unramified extension. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves finite ramification support for radical extensions.

The local argument does not assume ${\mu }_n\subset K$: adjoining all roots proves the stronger unramified assertion for the splitting field, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.