Theorem 2.9-1 in Functional Analysis book of Kreyszig 
My question is that in Eq. (7) $x$ can be any element of $X$ but why afterwards it evaluates the problem only for those $x=e_j$?
At fisrt it seemed that if it is true for basis ${\{e_j}\}$ then it will be true any $x$ but I am wrong becasue $$0= \sum_{k=1}^n \beta_k f_k(x) = \sum_{k=1}^n \beta_k \sum_{j=1}^n f_k(\xi_j e_j) =  \sum_{k=1}^n \beta_k \xi_k, $$ which doesn't imply each $\beta_k$ is zero.
 A: He's first trying to show that $F$ is a linearly independent set.  What is the definition of $F$ being a linearly independent set?  It means that no element of $F$ can be written as a linear combination of the others.  The expression
$$
\sum^n_{k=1}\beta_kf_k=0
$$
represents all possible ways that an element of $F$ can be written as a linear combination of other elements of $F$.  To see this for $f_i$ just choose $\beta_i=1$ and rearrange to get:
$$
f_i=\sum_{k\neq i}\beta_kf_k.
$$
But by even just evaluating the first expression at $x=e_j\in X$ we see that $\beta_j=0$.  For the case of $f_i$, choose $j\neq i$ to get:
$$
0=f_i(e_j)=\Big(\sum_{k\neq i}\beta_kf_k\Big)(e_j)=\beta_j.
$$
So there are no nonzero choices of the coefficients $b_j$ for $j=1,...,n$ that result in the first expression holding, which means that no element of $F$ can be written as a linear combination of any others.
EDIT:  So an easier way to show that $F$ is a linearly independent set than in Kreyszig's proof is simply to choose $i$ arbitrary and suppose otherwise, that we have 
$$
f_i=\sum_{k\neq i}\beta_kf_k.
$$
for at least one nonzero $\beta_k$ for $k\neq i$.  But then evaluating $f_i$ at any $e_j$ for $j\neq i$ results in $\beta_j=0$, a contradiction.
EDIt:  The response by anonymous is correct but then you need to show that $F$ being independent is equivalent to the statement that
$$
\sum^n_{k=1}\beta_kf_k=0\qquad\Rightarrow\qquad\beta_k=0\quad\forall k=1,...,n.
$$
A: $F$ being linearly independent is equivalent to the statement that $a_1 f_1 + ... + a_n f_n = 0$ iff all of the $a_i$ are $0$.  In this case, the $f_i$ are functions from $X \to \mathbb{R}$ defined by $f_i(e_j) = \delta_{i,j}$.
His argument proceeds as follows:
if $0 = \sum_j \beta_j f_j$ (where this is interpreted as the function $(\sum_j \beta_j f_j)(x) = \sum_j \beta_j f_k(x)$), then we have for every $x \in X$ that
$0 = \sum_j \beta_j f_j(x)$.  Choose $x = e_i$ for some index $i.  Then this sum is simply
$\sum_j \beta_j f_j(e_i)$.  For $j \neq i$ we have that $f_j(e_i) = 0$ and $f_i(e_i) = 1$.  So this reduces to
$0 = \beta_i$.
But $i$ is entirely arbitrary so we must have $\beta_i = 0$ for all $i$.
A: Equation (7) holds for every $x\in X$. His argument is that the   function $g=\sum_{k=1}^n \beta_k f_k$  is an element of $X^*$. If $g$ was the zero element of $X^*$, that is if $g(x)=0$  for every $x\in X$, then $g(e_k)=0$ for every $k=1, \ldots, n$ as well.
