why are subextensions of Galois extensions also Galois? Generally a Galois extension is defined to be an algebraic extension that is also normal & separable. It is then shown that in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This follows since the same property is valid for separable & normal extensions individually. It also follows that $L|K$ is a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$
In Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory he takes the opposite tack, and in definition 3.2.1.1 defines an algebraic extension of fields $L|K$ to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$.
It is then shown that in the sequence of algebraic field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction (and so has no details), I don't see the obviousness...can someone clarify.
In proposition 3.2.1.3 he shows that Galois extension is normal and separable.
All this appears to be in the opposite order of the standard treatments. One reason I'm interested in his formulation, if it is correct, is that one side of the Galois correspondence follows easily from this.
disclaimer: I've already asked this question on math.stackexchange but the answers there revolved around characterising Galois extensions as being normal & separable, and then showing this property follows.
 A: This proof might not count as direct. It is for finite extensions.
Lemma: $L/K$, a finite extension, is Galois if and only if $L \otimes_K L$ is a product of copies of $L$.
Proof: If $L\otimes_K L$ is a product of copies of $L$, and $x$ is fixed by every automorphism, then $s \otimes 1- 1 \otimes s$ is zero in $L \otimes_K L$, so $s \in K$ (via explicit description of tensor products of vector spaces.) 
If there are a lot of automorphisms, then each automorphism gives a different surjective map $L \otimes_K L \to L$, so we get a surjective map from $L \otimes_K L $ to the product of $[L:K]$ copies of $L$, which must be an isomorphism by dimension-counting. 
Then $L \otimes_M L$ is a quotient of $L \otimes_K L$, so is a product of finitely many copies of $L$.
I believe one can extend this to all algebraic extensions via a slightly more complicated argument. But that might just be pointless, as one could argue that my condition is just a clever way of saying "normal and separable" in different language.
A: Edited (in view of new comments to this answer and the original question): I believe that Delgado missed the point that $M=Fix(Aut_M(L))$ isn't a formal consequence of $K=Fix(Aut_K(L))$ for algebraic extensions $K\subseteq M\subseteq L$. I took a closer look into the (master?) thesis. It doesn't claim to contain anything new, This work is a journey through the main ideas and sucessive [sic] generalizations of Galois Theory, towards
the origins of Grothendieck’s theory of Dessins d’Enfants ... as the author puts it in his abstract.
The chapter on Galois theory just repeats well-known text book material, mostly without proofs. Considering the verbose character of this chapter, I'm sure the author would have said more than We immediately conclude that ... if there had been a novel aspect. To me it appears that he simply missed an essential aspect of Galois theory.
At any rate, from $K=Fix(Aut_K(L))$ alone we cannot conclude much, one somehow has to use the fact that $L/K$ is algebraic too as the following example shows: If $L=K(x)$ for a transcendental $x$, and if $K$ is infinite, then $K$ is the fixed field of $Aut_K(L)$, but for most rational functions $r(x)$ the extension $L/K(r(x))$ isn't Galois in either sense.
So if we want to show that $M=Fix(Aut_M(L))$ for an algebraic extension $L/K$ with $K=Fix(Aut_K(L))$, then I believe that one is automatically lead to the usual kind of arguments, which by the are also listed in this thesis. 
