I know that separable extensions form distinguished class. The property which is used here is the following: Suppose we have field extensions $k<F<E$. Then $E/k$ is separable if and only if $F/k$ and $E/F$ are separable.
But in proposition 6.6. we are considering all subfields $F$ such $k<F<E$ such $F/k$ is separable.
If the number of such subfields is finite then it's OK.
But what if we have infinite number of such subfields? How do we have to use the property of distinguished class?
Using only the distinguishedness, this is not possible. For example, finite extensions are a distinguished class, but the compositum of, say, all subfields of $\Bbb C$ that are finite extensions of $\Bbb Q$ is not a finite extension of $\Bbb Q$.