Can I always order a countable set of numbers

" Question: I CAN do that right?"

Actually, no. The "usual" order (where $a < b$ means $b-a$ is positive) is not what is called a "well-ordering", i.e. an order in which there is always a "least" element of all subsets.

You have proven that "$<$" is not a well-ordering as $\mathbb Q \cap (0,1)$ does not have a least or greatest element. (Neither does $\mathbb Z$ or $\mathbb Q$).

!!!!BUT!!!!!, a "well-ordering" of a countable set will always exist! It just won't be "$>$".

You have actually proven that. Let $x \prec y$ mean that $f^{-1}(x) < f^{-1}(y)$. Then "$\prec$" is a well ordering and the "least" element is $x_1$.

"$\prec$" has all the properties of an "order". (Exactly one of $a \prec b$ or $b \prec a$ or $a = b$ are true; $a \prec b$ and $b \prec c \implies a \prec c$). But it doesn't mean what we think of as "bigger" and "smaller". It means nothing more or less than "comes before in a list". Which is just as valid a basis for order and "being smaller" is.

Post-script (and maybe more to the point):

"It seems right, since any xi and xj are different, then one is bigger than the other, thus I can do this comparison with all of them and order them."

Ah!... good question. The problem with this is that you never finish. You can always say "I've sorted 5 billion of them and so far $x_{1,672,453,928}$ is least and $x_{17}$ is the greatest" but then you will also have to admit "but I can not be certain that they will remain the largest and smallest... And I'll never be certain."

Try it with $\mathbb Z$". We can list all the integers in order as $0, 1,-1,2,-2, 3,-3....$.

So "$1$ is bigger than $0$ so keep $1$. $1$ is bigger than $-1$ so keep $1$. $2$ is bigger than $1$ so replace $1$ with $2$. $2$ is bigger than $-2$ so keep $2$. $3$ is biggr than $2$ so replace $2$ with $3$. $3$ is bigger tan $-3$ so..."

You see how that will never actually end and we will never actually get a largest or a smallest element.

Post-Post-Script: Notice a well ordering says all subsets will have a least element. It doesn't actually say it will have a largest element. And "comes earlier in a list" makes sets that do always have a first element but do not always have a last element.

It should be intuitive though that any axiom of definition we make for a well-ordering always having a least element could arbitrarily be make for a "perverse" ordering which always have a greatest element but not nescessarily a least. If "$\prec$" is a well-ordering, then defining "$\prec_{ident}$" as $a \prec_{ident} b \iff b \prec a$ will, of course, be a perverse ordering.

("perverse ordering" and "$\prec_{ident}$ are terms I made up and not valid mathematical terminology. I hope that was clear.)

(That's probably a lot more than you asked for.)

$X = \mathbb{Q} \cap (0, 1)$ is a countable set. An explicit enumeration of this set is $$x_1 = 1/2, \quad x_2 = 1/3, \quad x_3 = 2/3, \quad x_4 = 1/4, \quad x_5 = 3/4, \quad \ldots$$ Therefore, we can construct a total order $<_X$ on the set $X$ as follows: given two numbers $x_i$ and $x_j$ in $X$, we write $x_i <_X x_j$ whenever $i < j$.

This does not correspond to the usual order $<$ on the real numbers, under which there is no smallest element in $X$. On the other hand $x_1 = 1/2$ is the smallest element in the order $<_X$.

Addendum: I realize after writing this that you asked for an order that has a largest element, but the argument is the same, defining instead $<_X$ by $x_i <_X x_j$ whenever $i > j$.

You always can do it. You are making a mistake when says which there is not biggest rational in $(0,1)$. What is the mistake? There ist not biggest rational in (0,1) using the standard order relation. This order that you are defining, is not the standard. See by well ordenation, every set can be well-ordered.