Questions about infinite dimensional normed spaces Problem
Show that if $X$ is an infinite-dimensional normed vector space, then there is a sequence $\{x_n\}$ such that $\|x_n\|=1,\|x_n-x_m\|=1\ (n\neq m)$．
My view
I have confirmed that the following two theorems hold.
Theorem1:If $M$ be a close subspace of a normed vector space $X$ which is $X\neq M$, then for all $\varepsilon>0$ there is $u_{\varepsilon}\in X$ such that $\|u_{\varepsilon}\|=1,d(u,M)>1-\varepsilon$．
Theorem2:By Theorem 1, if $M$ is finite dimensional, we can get $u_0\in X$ such that $\|u_0\|=1,d(u_0,M)\geq1$.
I think I will use these theorems, but I can't think of a specific way to do it.
Also, from a geometric point of view, if $X$ is one-dimensional (line), we can get $x_1\in X$ where $\|x_1\|=1$.
If $X$ is two-dimensional (plane), then by drawing an equilateral triangle with one point at the origin, we can obtain $x_1,x_2\in X$ such that $\|x_i\|=1\ (i=1,2),\|x_1-x_2\|=1$.
Furthermore, if $X$ is three-dimensional (space), by considering a regular square pyramid with a single point at the origin, we can obtain $x_1,x_2,x_3\in X$ such that $\|x_i\|=1\ (i=1,2,3),\|x_i-x_j\|=1\ (i\neq j)$.
I was wondering if I could do the same thing in higher dimensions, so that I could do the same thing in infinite dimensional space, but it didn't work.
Postscript (Nov. 11, 2021)
I still haven't solved this problem yet, so I'll describe another method that I've newly thought of.
My view (New)
Take $x_1\in X$ such that $\|x_1\|=1$, and let $M_1$ be a finite-dimensional subspace of $X$ spaned by $x_1$.
From Theorem 2, we obtain $x_2\in X$ such that $\|x_2\|=1$ and $d(x_2,M_1)\geq 1$.
Next, let $M_2$ be a finite dimensional subspace of $X$ spaned by $x_1$ and $x_2$. From Theorem 2, we obtain $x_3\in X$ such that $\|x_3\|=1$ and $d(x_3,M_2)\geq 1$.
Continuing this work, let $M_n$ be a finite dimensional subspace of $X$ spaned by $x_1,\cdots,x_n$. From Theorem 2, we obtain $X_{n+1}\in X$ such that $\|x_{n+1}\|=1$ and $d(x_{n+1},M_n)\geq 1$.
Assume that this task is completed in a finite number of times. In this case, $X$ is a finite dimensional space, which is a contradiction. Therefore, this work can be done infinitely many times.
In this case, $\{x_n\}_{n=1}^{\infty}$ satisfies $\|x_n\|=1\ (n=1,2,\cdots)$, and from $d(x_{n+1},M_n)\geq 1$, we can say $1\leq\|x_n-x_m\|$ for $n\neq m$.
My Question
Now if I can show that $\|x_n-x_m\|\leq 1\ (n\neq m)$, I think the proof is done.
Do you have any better ideas?
 A: If you construct a sequence in the way you've described in your postscript, there is no reason that it would have to be be the case that $||x_n - x_m|| = 1$ for all $m \neq n$.  In fact, most such sequences do not have this property.
Since you're having trouble with the logical setup of the proof, let's actually specify in detail how you could logically establish that such a sequence exists.

First of all, it's going to be easier to think about sets  than sequences here.  The ordering isn't really important, so it's enough to figure out what's happening for subsets $S \subset X$ of unit vectors such that
\begin{equation}
||s|| = ||s - t|| = 1 \text{ for each } s \neq t \in S \qquad (\dagger)
\end{equation}
Sets satisfying $(\dagger)$ certainly exist: any singleton $\{x\}$ where $x$ is a unit vector is an example, as is the empty set.  Moreover, if I have an ascending chain $S_1 \subset S_2 \subset \cdots$ of sets satisfying $(\dagger)$ then the union $S = \bigcup S_i$ of this chain satisfies $(\dagger)$.  So it follows by Zorn's lemma that there exist maximal sets satisfying condition $(\dagger)$.
Take a maximal set satisfying  $(\dagger)$.  This set must either be finite or infinite.  If it's infinite, then we're done, because all we wanted was to show that there was at least one infinite set satisfying $(\dagger)$.  (Actually we wanted an infinite sequence, but it's easy enough to produce an infinite sequence from an infinite set by just picking elements arbitrarily.)
So we need to show that a finite set satisfying $(\dagger)$ cannot be maximal among all sets satisfying $(\dagger)$.  Unpacking this, what it says is:

Suppose we have a finite subset $\{x_1, \ldots, x_n\} \subset X$ such that $||x_i|| = ||x_i - x_j|| = 1$ for all $i \neq j$.  We wish to show that there exists $y \in X$ with $||y|| = ||y - x_i|| = 1$ for all $i$.


Note that everything we've said above has just been establishing the logical structure of a proof.  We haven't done anything to do with normed vector spaces at all yet.  Since that's the subject of your class, I'll leave that part up to you.
(By the way, some people might worry whether the inclusion of Zorn's lemma above might be overkill.  I suspect not, because we need Zorn's lemma for closely related things like proving that every vector space has a basis.  We can presumably get along with the weaker version of Zorn's lemma equivalent to countable choice here, since we are only trying to find infinite sequences, as opposed to infinte sets spanning $X$.
Also, I think students may sometimes get a misleading impression that whether we're using the axiom of choice is something that's important to worry about all the time.  It's not, and in fact I suspect that maybe 50% of Fields medalists probably couldn't tell you the ZF axioms off the tops of their heads.  Foundations are roughly as important to mathematics as electricity and magnetism are to computer science, which is to say that they're not important at all unless you're the 1 person out of 100 who happens to do work that deals with them.)
