I believe the answer is `No', and Franz Lemmermeyer's example $K=Q(2^{1/3})$ and his strategy of the proof do the trick.

Suppose this particular $K$ is contained in the compositum $F$ of $L_i$, with every $L_i$ ramified at only one prime. Assume each $L_i$ is Galois over $Q$ (otherwise replace it by its Galois closure) and that no two $L_i$ ramify at the same prime $p$ (otherwise replace this pair by their compositum). The $L_i$ are then linearly disjoint over $Q$, since their pairwise intersections would have to be unramified at all primes. So $G=Gal(F/Q)$ is the direct product of $Gal(L_i/Q)$'s.

Now the group $G$ has a 2-dimensional irreducible representation $\rho$, the one that factors through the Galois closure of $K/Q$ (an $S_3$-extension of $Q$). As $G$ is the direct product of groups, we can write $\rho=\rho_1\otimes...\otimes\rho_n$ uniquely, with $\rho_i$ irreducible representations of $Gal(L_i/Q)$. Moreover, $\rho$ is self-dual, so all the $\rho_i$ are self-dual as well. Of these $\rho_i$ one must be 2-dimensional and the others are 1-dimensional. Because 1-dimensional self-dual characters have order 2, this shows that at all primes $p$ except at most one (the one corresponding to the 2-dimensional $\rho_i$) inertia $I_p$ acts on $\rho$ through a quotient of order 2. But $I_2$ and $I_3$ act through quotients of order 3 and 6 respectively, contradiction!

over $\mathbb{Q}$. $\endgroup$ – Alex B. Nov 13 '10 at 11:41