I am looking for help understanding the proof that $\mid Gal(E/F)\mid\leq [E:F]$ by induction on the degree of the extension. $E/F$ is a finite algebraic extension. I'll mark them by [1], [2], [3]. I just have specific questions on specific parts. This is my attempt at phrasing it.
Note: This does not assume separability or normality. I think some authors use $Aut$ instead of $Gal$ and reserve the notion of a galois group when the extension is a Galois one, but I think in my course we just always said galois group.
Define $Gal(E/F):=\{\phi:E\rightarrow E \text{ isomorphism}\mid \phi|_{F}=\text{id}_F\}$
Proof
Firstly if $[E:F]=1$ then we can take $E=F$ and the galois group is just the identity automorphism of $F$.
Now suppose $[E:F]>1$ and the induction assumption holds less than this degree.
Let $\alpha \in E$ algebraic over $F$ and $m$ it's minimal poly. over $F$ with degree $n$ and suppose there are $r \leq n$ roots of $m$ in $E, \;\alpha_i, \; i \in \{1,\ldots, r\}$.
If $E=F(\alpha)$, then for $\phi \in Gal(E/F), \; \phi$ is completely determined by $\phi(\alpha)$ [1]
There are at most $n$ choices so $\mid Gal(E/F) \mid \leq n =[E/F]$.
Now suppose that $F<F(\alpha)<E$. Consider $H:=Gal(E/F(\alpha))$. By induction $\mid H \mid\leq [E:F(\alpha)]$ and $H\leq Gal(E/F)$.
Let $\phi \in Gal(E/F)$ and let $\phi(\alpha)=\alpha_i$.
Suppose there exist $\phi_i \in Gal(E/F)$ s.t. $\phi_i(\alpha)=\alpha_i$ for $i \in \{1,\ldots,r\}$, [2]
then $\phi \circ \phi_i^{-1} (\alpha)=\alpha$ so $\phi \circ \phi^{-1}_i \in H$ and hence $\phi \in H \phi_i$.
Then $Gal(E/F)\subseteq \bigcup\limits_{i=1}^r H \phi_i$. [3]
Thus $\mid Gal(E/F)\mid\;\leq \;\mid H\mid\cdot r \;\leq [E:F(\alpha)] \cdot[F(\alpha):F]=[E:F]$.
[1] Is this simply because the extension is determined in terms of a basis of powers of $\alpha$ over $F$?
[2] Should this $r$ be something like $t\leq r$ which is mentioned at the start. When we were doing the proof it was pointed out that not all the roots may be able to be mapped to. I couldn't really grasp that at the time, and possibly have taken that down wrong. Eg looking at $\mathbb{Q}(\sqrt[4]{2})$ and we cannot have an isomorphism of this field taking $\sqrt 2 \rightarrow -\sqrt 2$. In which case the $r$'s afterward should be this new $t$.
[3] Is this because each automorphism in the galois group has to map $\alpha$ to some $\alpha_i \in E$ in which case as shown it is also in the coset $H \phi_i$?
If $\alpha$ is algebraic over $F$, then $F(\alpha)$ equals $F[\alpha]$. So every element of $F(\alpha)$ is of the form $f(\alpha)$ for some $f(x) \in F[x]$. Thus, if $\phi \in Gal(F(a)/F)$ then $\phi(f(\alpha)) = f(\phi(\alpha))$. So you are right at [1].
For [2], I think your idea could be better formulated as: if $\phi, \psi \in Gal(E/F)$ and $\phi(\alpha) = \psi(\alpha)$ then $\phi \equiv \psi \mod H$.
For [3], if $\phi \in Gal(E/F)$ then certainly $\phi$ sends a root of a polynomial in $F[x]$ to another root. So there are at most $r$ many cosets.