Density of set of splitting primes Let $K$ be a number field and let $S$ be a set of primes of $K$ containing the set of archimedian primes $S_\infty$. Suppose, $S$ has Dirichlet density $\delta(S) = 1$. 
Then the claim is that the set of completely splitting primes in the extension $K(\mu_{p^r}) | K$ for every rational prime $p$ and $r \in \mathbb{N}$ arbitrary is contained in $S$ up to a set of Dirichlet density $0$.
My guess is to show, that the Dirichlet density of this set of completely splitting primes is $1$ as well, but I don't know how to apply Chebotarev's density theorem.
Thank you!

Let $P$ be the set of all primes of $K$. If $S\subseteq P$ has full Dirichlet density $\delta(S)=1$ then we see its complement has density $\delta(P\setminus S)=0$ and hence every subset of $P\setminus S$ has density $0$. Thus if we have any subset $X\subseteq P$, the difference $X\setminus S$ has density $0$, i.e. $X\subseteq S$ up to a set of density $0$.
In your case you can set $X$ to be the set of primes that completely split in all cyclotomic extensions by prime power roots of unity, but these hypotheses on $X$ aren't really needed to see the fact.
