Here is a solution using some representation theory of the quaternions and Dirichlet's theorem on the units:

the quadratic subfields of our quaternion extension (call it $F$) are of the form $\mathbb{Q}[\sqrt{a}]$, $\mathbb{Q}[\sqrt{b}]$, $\mathbb{Q}[\sqrt{ab}]$, corresponding to, say, $i$,$j$,$k$, respectively, in $Q_8$. So either all three are real or exactly two are complex and one is real. Assume, the latter is the case and wlog $a,b<0$. Then in the following table I list the intermediate fields, the number of real embeddings $r_1$, the number of pairs of complex embeddings $r_2$ and the rank of the unit group, which is $r_1+r_2 -1$:

- $F$: 0, 4, 3
- $\mathbb{Q}[\sqrt{a},\sqrt{b}]$: 0, 2, 1
- $\mathbb{Q}[\sqrt{a}]$: 0, 1, 0
- $\mathbb{Q}[\sqrt{b}]$: 0, 1, 0
- $\mathbb{Q}[\sqrt{ab}]$: 2, 0, 1
- $\mathbb{Q}$: 1, 0, 0

The units of $F$ (thought of as an abelian group) tensored with $\mathbb{Q}$ therefore yield a three dimensional rational representation of our Galois group $Q_8$, call it $V$. Now, the irreducible rational representations of $Q_8$ are the trivial representation, three sign representations and a 4-dimensional one, namely twice the complex irreducible 2-dimensional representation. So $V$ is a sum of a certain number of trivial and sign representations, 3 altogether. Now, there can be no copies of the trivial representation in $V$, since the rank of the units over $\mathbb{Q}$ is 0. But by the same argument, there can be no copy of the sign representations that factor through $i$ and $j$, so the only one occuring in $V$ is the one factoring through $k$. But then the rank of the fixed subspace of $V$ under $k$ would have to be the same as the rank of $V$ itself, which is a contradiction.

Note that to write out all the details was slightly lenghty, but the idea is simple: look at the rank of the units in all the subfields and show that this cannot correspond to a rational representation of $Q_8$.