why do I have to restrict the measurement standard to real number? Why not any ordered field?
You can use any ordered field, and the axioms will still make sense. The thing is, though, that most of our geometric intuition is built on the Archimedian property. But if we include this in our field, then it becomes a subfield of $\Bbb R$. In this case, we are really not restricting ourselves to $\Bbb R$ at all: it's the largest Archimdean ordered field!
Thinking about ordered fields that aren't Archimedian gets a little more weird. Are you prepared, for example, to have vectors $v,w$ such that $v$ and $w$ point in the same direction, and yet $vn$ is shorter than $w$ for every natural number $n$?
I mean, if we define $\|v\|=\langle v, v\rangle^{1/2}$ then indeed $(V, \|\, \|)$ becomes a normed vector space
No, not quite! Normally we want the norms to be in the base field, but the vector $(1,1)\in \Bbb Q^2$ would have a norm outside of $\Bbb Q$, with that definition. To fix things, you'd need something called a Pythagorean field, and that's enough to guarantee this definition of norm works.
If you don't ask for the norm to be in the base field, then you may not be able to carry out normalization, because dividing by the norm won't give you a vector in the field.
I know that the real number is the only (up to isomorphism) ordered field with the least upper bound property and the property that every increasing bounded sequence converges. But is that enough to justify that every metric spaces use real number? I want to be more convinced, then.
Yes, your intuition is along the right lines. The fact that $\Bbb R$ is "maximal" among Archimedian ordered fields makes it special. It's a smooth connected piece with no holes. In geometry this is important since it ensures that lines and circles cross where you expect them to. For example in $\Bbb Q^2$, you can find an example of a line and a circle that would intersect in $\Bbb R^2$, but they pass through each other in $\Bbb Q^2$ without touching.
Why not... anything else!!
Actually, geometers study generalizations of the norm in the form of bilinear forms over any field. The idea is that rather than focus on the norm, you instead focus on a (generalized) inner product. They can be quite different from what you're used to with run-of-the-mill normed real spaces. They can have, for example, nonzero vectors with length $0$ or even negative length.
Even more than that, "length" loses meaning totally when you're working in a field that isn't ordered: there's no such thing as positive or negative, there. Still, there is a huge theory for these types of spaces with (generalized) inner products.
Best Answer
Suppose $x$ is a positive real number. It can be shown that $n\mapsto 10^n$ is a map that is unbounded above in the integers, so by Archimedean property, there is some integer $n$ with $10^{n+1}\ge x$. Take the least such $n$ (why must one exist?), and let $a_{-n}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $a\cdot 10^n<x$ (why must one exist?). Now, given $a_{-n},a_{-n+1},...,a_{m}$ for some $m\in\Bbb Z$ with $m\ge-n,$ we let $a_{m+1}$ be the greatest element $a$ of $\{0,1,2,...,8,9\}$ such that $$a\cdot 10^{-(m+1)}<x-\sum_{j=-m}^na_{-j}\cdot10^j$$ (why must one exist?) Recursively, this determines a sequence $a_{-n},a_{-n+1},...$ of elements of $\{0,1,2,...,8,9\}$ such that for all integers $m\ge-n$ we have $$S_m:=\sum_{j=-m}^na_{-j}\cdot 10^j<x.$$ In fact, $S_{-n},S_{-n+1},...,S_m,...$ is a non-decreasing sequence of positive numbers, so since bounded above by $x,$ this sequence converges to some number no greater than $x$ by the Monotone Convergence Theorem. We can even do better than that, and show that the sequence of partial sums $S_n$ converges to $x$ (why?). The series thus determined is the infinite decimal expansion of $x$.
If $x$ had been negative, we could acquire a decimal expansion for $-x$ in this way, and then the opposite of that would be a decimal expansion for $x$. $0$ has a decimal expansion, too, so all real numbers have a decimal expansion, and moreover, all of them (except arguably $0$) have an infinite decimal expansion (though some also have a finite decimal expansion).