>To prove the statement" X is an infinite-dimensional normed vector space then X is not locally compact". I read the proof of Must a complete space be locally compact? by @martini. But I still have some questions.

To prove the statement" X is an infinite-dimensional normed vector space then X is not locally compact".

In his proof, he did not claim the sequence with norm 1. I read a proof:
Suppose that $0 \in X$ had a compact neighborhood $C$, and choose $r>0$ such that $B_{2r}(0) \subset C$. Since scalar multiplication is continuous, $(1/r) C$ is compact and contains all $x \in X $ with $\Vert x \Vert = 1$. Then we can find a sequence $x_j$ with $\Vert x_j\Vert=1$ and $\Vert x_j-x_k\Vert>1/2$ in $(1/r) C$ which has no convergent subsequence(every ball of radius $1/4$
contains at most one point from the sequence). This is a contradiction.

Q1: Why "$(1/r) C$ is compact and contains all $x \in X $ with $\Vert x \Vert = 1$"?
Q2: What does "(every ball of radius $1/4$ contains at most one point from the sequence)" mean?

 A: Q1: The function $g:X \to X$ defined by $g(x)=\frac 1 r x$ is a continuous map. Hence $g(C)$ is compact. Note that $g(C)=\frac  1 r C$. If $\|x\|=1$ then $\|rx\|=r <2r$ so $rx \in B_{2r}(0) \subseteq C$. Hence $x \in \frac 1 r C$. 
Q2: Suppose a ball $B(y,\frac  1 4)$ contains $x_i$ and $x_j$ with $ i \neq j$. Then $\|x_i-x_j\|\leq \|x_i-x\|+\|x-x_j\|<\frac 1 4 +\frac 1 4 =\frac 1 2$ which is a contradiction. 
A: This result is true   for  topological vector spaces without any norm  assumption. Please see page 17 theorem 1.22  of Functional  Analysis, Walter Rudin.
If  I  am  not  mistaken this  apparently simple  theorem is essentially used in the  process of  index theorem(Atiyah Singer Index theorem).  More  precisely the  Kernel of  an elliptic differential operator is  a  finite dimensional space.  I remembere I learned it from a Lecture  by  Nicolaescu. He arrived at a point proving the kernel is  a  locally compact space  hence is  finite dimensional space.
Edit: I just find it: See page  58 of this link.
I quote the corresponding part of this page "We will now show that any ball in ker E which is closed with respect to the $L^2$-norm must be compact in the topology of this norm. The desired conclusion will then follow from a classical result of F. Riesz, [6, Ch. VI] according to which a Banach space is finite dimensional if and only if it is locally compact"
