$K' \subseteq K \Leftrightarrow Desc(K) \subseteq Desc(K')$ Let be $K/Q$ a normal extension and define $Desc(K) = \{ p \in Q: p$ splits completely in $K \}$. Show that 
$K' \subseteq K \Leftrightarrow Desc(K) \subseteq Desc(K')$ 
I could do the first part. The idea is, if $K' \subseteq K$, and $p \in Desc(K)$, then we have $(p) = \wp_1 ... \wp_g$ and $e(\wp_i | p ) = 1$ for all $i$. Taking $ \wp_i' = K' \cap \wp_i $ and using that $1 = e(\wp_i | p ) = e(\wp_i | \wp_i')e(\wp_i' | p )$ we see easily that $e(\wp_i' | p ) = 1$. In a similar way, $f(\wp_i' | p ) = 1$.
But I don't have idea how to do to prove the other side.
 A: Suppose that every prime which splits completely in $K$, also splits completely in $K'$.  To show that $K' \subset K$, it suffices to show that $K = K'K$.
Of course, every prime which splits in $K'K$ also splits in $K$.  
Lemma: Under our hypothesis, the primes which split in  $K$ are the same as the primes which split in $K'K$.
Proof: Of course every prime which splits in $K'K$ also splits in $K$.  The converse follows from our hypothesis and the general fact that if $p$ splits in $K'$ and $K$, then it also splits in $K'K$.   $\square$
It follows from the lemma that what you're asking is reduced to the following proposition:
Proposition: Let $F \subset E$ be Galois extensions of $\mathbb Q$.  If the same primes which split in $E$ also split in $F$, then $F = E$.
The proposition follows from the Chebotarev density theorem, which states the primes which split completely in $F$ has density $\frac{1}{[F : \mathbb Q]}$.  Applying the same principle with $E$, you get $[F : \mathbb Q] = [E : \mathbb Q]$, hence $E = F$.
