i think my proof makes sense but i'm unsure of quality.
$e_j$ is a vector whose jth coordinate is 1 and all other coordinates are 0. all vectors have n elements.
let $a_i \in F, i=0,...,n$
on the left side of the following results, $0$ denotes the zero vector.
the linear combination of all elements in the set is:
$0=a_1e_1+...+a_ne_n$
$0=a_1(1,0,...,0,0)+...+a_n(0,0,...,0,1)$
$0=(a_1,0,...,0,0)+...+(0,0,...,0,a_n)$
$0=(a_1,...,a_n)$
the only way for the above identity to hold is if $a_i=0$.
therefore {$e_1$,...,$e_n$} is linearly independent.
Yes, your proof is correct.
Since we are in $F^n$, $0$ is defined to be the element $(0,\dots, 0)$, so from the linear combination we have a system of equations of the form $a_i=0$.
Note that in general, expressing a vector as $(a_1,\dots,a_n)$ depends upon choosing a basis and saying that those are the coordinates with respect to that basis, in particular that is true for $0$. What I mean is that if you don't have a basis (more particularly, if your vectors are not linearly independent), $0$ could be written in different ways, so the fact that it can be written as $(0,\dots, 0)$ doesn't imply that that's the only way of writing it.
So the caveat for a more general case is: don't use linear independency to show linearly independency.