Okay, I think I finally get it. I think my confusion came from the quantification on $x \in X$. Please correct me if I'm still not getting it!
We are assuming $\sum_k^n \beta_k f_k(x)=0$ for every $x \in X$ (since $X$ is the domain of $f_k$).
Now, we choose $x = e_k \in X$, and this leads us to...
Thanks for the post.
I'm still not convinced I guess. Here's another way to ask my question:
Let's say $x = \sum_j^n \xi_j e_j$ and then act with $(*)$ on $x$. Assume some of the $\xi_j \not = 1$.
So we assume $\sum_k^n \beta_k f_k(x) = 0$.
$f_k(x) = \sum_j^n \xi_j f_k(e_j) = \xi_k$
Then...
This is from Kreyszig's Introductory Functional Analysis Theorem 2.9-1.
Let $X$ be an n-dimensional vector space and $E=\{e_1, \cdots, e_n \}$ a basis for $X$. Then $F = \{f_1, \cdots, f_n\}$ given by (6) is a basis for the algebraic dual $X^*$ of $X$, and $\text{dim}X^* = \text{dim}X=n$...