Prove that $(0 ,1)\cap\mathbb{Q}$ is order-isomorphc to $\mathbb{Q}$: is my proof correct? I must prove that $(0 ,1)\cap\mathbb{Q} \approx \mathbb{Q}$. Is my proof below correct?
(Note: I know the 'every two linearly ordered, dense, countable sets with no minumum and no maximum are order-isomorphic' theorem, but my textbook suggests to do it this way. However, any alternative proofs are very welcome).
My proof
First, let's map $ \frac{1}{2} \in (0 ,1)\cap\mathbb{Q}$ to $0 \in \mathbb{Q}$.
Now, let's consider sequence $\frac{n}{n+1}$, where $n = 2, 3, 4, \dots$ (the sequence converges to $1$), and map its every element to a positive integer in $\mathbb{Q}$, like this:
$\frac{2}{3} \mapsto 1$,
$\frac{3}{4} \mapsto 2$,
$\dots,$
$\frac{n}{n+1} \mapsto n-1$.
And let's consider sequence $\frac{1}{2^n}$,
 where $n=2, 3, 4, \dots$ (the sequence converges to $0$), and map its every element to a negative integer in $\mathbb{Q}$, like this:
$\frac{1}{4} \mapsto -1$, 
$\frac{1}{8} \mapsto -2$, 
$\dots$, 
$\frac{1}{2^n} \mapsto -n+1$.
We have 'an order-isomorphism template':

Consider two dots: $(f, z)$ and $(f', z')$, such that $f$ and $f'$ are either $\frac{n}{n+1}$ and $\frac{n+1}{n+2}$, or $\frac{1}{2^{n+1}}$ and $\frac{1}{2^n}$, and $z$ and $z'$ are corresponding integers.
$f$, $f'$, $z$, and $z'$ are all rational, so if we build a linear function by solving the system of equations
$$\begin{cases}
z=kf + b \\ 
z'= kf' + b, \\ 
\end{cases}$$
then $k$ and $b$ will also turn out to be rational.
And, as long as we input rational $x \in (0,1)\cap\mathbb{Q}$ into the function, the output $y \in \mathbb{Q}$ will be rational.

Example: $(f, z) = (\frac{1}{2}, 0)$, $(f', z') = (\frac{2}{3}, 1)$.
  From solving  $$ \begin{cases} 0 = \frac{1}{2}k + b\\ 1 = \frac{2}{3}k
 + b\\ \end{cases} $$ we get $y = 6x -3$.

By doing this to every two 'neighboring' dots, we get the required order-isomorphism:

 A: Your proof works, but there is one missing bit.
You have constructed a function $f:(0,1)\cap\mathbb Q\to\mathbb Q$ which is increasing and well-defined (really maps rationals to rationals).
What you didn't check is that it is surjective.
Are all rational values really achieved?
They are, but you should make that point explicitly.
If you do it interval by interval and look at your picture, surjectivity is pretty clear.
Whether it counts as a rigorous proof is a matter of taste.
In this case you could also do something more concrete, but the best choice is always a matter of taste.
You can actually give a formula for the isomorphism and its inverse.
That makes everything easy to check.
Perhaps the easiest starting point is that $f(x)=\frac{x}{1-x}$ gives an increasing bijection $[0,1)\to[0,\infty)$.
Since both $f$ and $f^{-1}(y)=\frac{y}{1+y}$ are rational functions, $f(x)$ is rational if and only if $x$ is rational.
It is important to use rational functions with both numerator and denominator of first order, since otherwise the inverse can fail to map rationals to rationals, making the function non-surjective between your two sets.
Now you can split your $(0,1)$ in two halves and stretch the right half to $+\infty$ and the left half to $-\infty$ with a variation of this $f$.
This leads to the function $g:(0,1)\to\mathbb R$
$$
g(x)
=
\begin{cases}
\frac{x-1/2}{1-x},&x\geq\frac12\\
\frac{x-1/2}{x},&x<\frac12
\end{cases}
$$
whose inverse is given by
$$
g^{-1}(x)
=
\begin{cases}
\frac{2y+1}{2+2y},&y\geq0\\
\frac{1}{2-2y},&y<0.
\end{cases}
$$
The basic shape of the plot is the same as yours.
The only difference seems to be how we engineered the function to map rationals to rationals.
