Find sequence of linear functionals s.t. $x_n^*(x_n)\not=0$ using Hahn-Banach I have the following claim.

CLAIM:
Let $X$ be an infinite dimensional Banach space. By tha Hahn-Banach theorem and the fact that $X$ is infinite dimensional, there are sequences $\{x_n\}$ in $X$ and $x_n^*$ in $X^*$  such that $x_n^*(x_n)\not=0$ and $x_n^*(x_m)=0$ for $n\not=m$.

This claim is stated on Lacey.
The Hanh-Banach theorem that I want to use is: given $S\subseteq X$ and $\varphi\in S^*$ there is an extension of $\varphi$ to $X^*$ with the same norm.
I tried to make $x_n^*(x_j)=\delta_{nj}$ but I don't think this is a bounded function... I also know how to prove the claim if I only have finite $x_n$'s... do you have any hints on this?
 A: If we had a Hilbert space, this would be easy to obtain by picking the $x_n$ all pairwise orthogonal. So, how do we mimic orthogonality in Banach spaces?
Firstly, start with a nonzero vector $x_1 \in X.$ Define, using Hahn-Banach, a functional of norm $1$ on $X$ called $\pi_1$ so that $\pi_1(x_1) = 1.$ Let $x_2$ be a nonzero element of the kernel of $\pi_1.$
Then, define $\pi_2$ similarly so that $\pi_2(x_1) = 0, \pi_2(x_2) = 1.$ Pick $x_3$ in the kernel of $\pi_2$ and $\pi_1.$ (Why must these kernels have nontrivial intersection? Well, here's a general lemma: If $f_1, ..., f_n$ are finitely many functionals on an infinite-dimensional vector space, there must be a nonzero element $x$ belonging to all of their kernels.
The proof is not so hard. If $0$ were the only element of all their kernels, then a vector $x$ in your space is uniquely determined by the $n$-tuple $(f_1(x), f_2(x), ..., f_n(x)).$ In this way, we get a linear injection into $\mathbb{R}^n,$ which is impossible by the fundamental theorem of linear algebra since $X$ is infinite dimensional.)
Likewise, define $\pi_1, ..., \pi_n$ in this way so that $\pi_i(x_j) = \delta_{ij}$ for $1 \leq i, j \leq n$ and pick $x_{n+1}$ a nonzero vector in the kernel of all the $\pi_1, ..., \pi_n.$ Define $\pi_{n+1}(x_{n+1}) = 1$ and $\pi_{n+1}(x_i) = 0$ for $i=1,2,..., n.$
It's easy to see why $x^*_n = \pi_n$ works in your problem, since we always pick in the kernel.
