Galois extension characterization Why to mention or why to give a basis for $E$ over $F?$ Why to make a relation between the basis and the minimal polynomial?
I think the proof is missing the 'therefore $E/F$ is Galois' at the end of the proof, anyway Why can we conclude that it's a Galois extension?

A finite extension $E/F$ is Galois if and only if $E/F$ is normal and separable.

Proof (just this side $\gets$):
Suppose $E/F$ is normal and separable.
If $w_1,w_2,...,w_n$ is a basis for $E$ over $F,$ let $f_i(x)$ be the minimal polynomial of $w_i.$
Then $E/F$ is a splitting field for the product of $f_i$ and each $f_i$ is separable because $E/F$ is separable. 
 A: The idea here is to use a previous theorem that if $E/F$ is the splitting field of a separable polynomial then $E/F$ is Galois. We know that $E/F$ is normal and separable.


*

*Normal means that if $\alpha \in E$ then all the conjugates of $\alpha$ are in $E$. If $f_{\alpha}(x)$ is the minimal polynomial of $\alpha$ then the roots of $f$ are the conjugates of $\alpha$ so $f_{\alpha}(x)$ has all of its roots in $E$ (it splits).

*Separable means that every minimal polynomial is a separable polynomial.
Thus for every element $\alpha \in E$ we get a separable polynomial $f_{\alpha}$ that splits. If there are infinitely many elements of $E$ then this doesn't help us so much. We use the finiteness condition to conclude that $E$ can be generated by a finite set $\alpha_1,\dots,\alpha_n$. That is, $E = F[\alpha_1, \dots, \alpha_n]$.
Now we take the corresponding minimal polynomials $f_1 = f_{\alpha_1}, \dots, f_n = f_{\alpha_n}$. We know that $E$ is the splitting field of $f_1,\dots,f_n$ because the splitting field contains $\alpha_1,\dots,\alpha_n$ and they generate $E$.
Finally the last trick is to note that if $E$ is the splitting field of $f_1,\dots,f_n$ then $E$ is the splitting field of the product $f := f_1 \cdots f_n$. The polynomial $f$ is separable because each irreducible factor $f_1,\dots,f_n$ is separable.
Therefore $E/F$ is the splitting field of a separable polynomial and we can apply the previous result.
