Proof for the existence of basic feasible solution I am trying to understand a proof for:

If $F = \{x \in \mathbb{R}^n: Ax=b, x \ge 0\}$ is non empty then it has a BFS.

The proof goes likes this, first we collect  all the indices j where $x_j > 0$, then we select the columns $A_j$. Now the columns can be linearly independent or linearly dependent. In case of linearly independent columns, the proof goes on to say that the actual basis of the matrix A is going to have more columns than $A_j$ and we can create an augmented matrix by adding columns from the technology matrix A and the resulting basis will definitely have linearly independent columns.
My lack of understand is here: $A_j$ should be the complete basis because if $x_j = 0$, then the solution is a degenerate one. My confusion is if there is no degeneracy, $A_j$ is the basis and how can we add more columns from A to create the augmented matrix?
Video of the proof: https://youtu.be/fEXLmsVn0Rw?t=2209
 A: Short answer: a degenerate solution is still a basic feasible solution.
In the video, lecturer proves that among all the feasible solutions in set $F$ there is at least one basic feasible solution. In order to do that, she chooses arbitrary feasible solution $x \in F$ — that solution does not have to be basic, and it certainly won't be when columns $\{A_j : j \in I
(x)\}$ are not linearly independent. But when they are, $x$ will be a basic feasible solution. As you correctly noted, if  $\{A_j : j \in I
(x)\}$ do not form a basis, it will be a degenerate solution, but it is nevertheless a basic feasible solution (as is shown in the video, by augmenting these columns with other columns from A, until they do form a basis).
So, if columns $\{A_j : j \in I(x)\}$ are linearly independent, there is nothing more to prove, and if they are not, the lecturer goes on, and uses $x$ to construct a basic feasible solution from $F$. 
In response to your comment: In the beginning of the proof, we pick some $x \in F$ and can't assume it has any other special property other than that it is a feasible solution, since we don't know yet if such $x$ exists in $F$. So we have to consider all the possibilities: the columns $\{A_j : j \in I(x)\}$ can be linearly independent, but they can also be linearly dependent. The first part of the proof deals with the former case, and second with the latter. 
Similarly, in the first case, when $\{A_j : j \in I(x)\}$ are linearly independent, we can't assume they form basis of the vector space spanned by the columns of $A$, but also have to consider the case when they do not, that is when there are less non-zero variables then $rank(A)$. Also, note that linear programs usually have many more variables than constrains and $rank(A)<n$ (where $n$ is number of variables), so when all the variables are non-zero, columns $\{A_j : j \in I(x)\}$ are certainly linearly dependent.
