First, I'll review my knowledge of 'what is $\aleph _1$'. My understanding is based on the video https://youtu.be/SrU9YDoXE88
So, say we list out the natural numbers in an infinite list. Now this set has cardinality $\aleph _0$ and order-type $\omega$. Then we take out every even number from the list and list them below our original list (the original list is only left with the odds now) Now this 'list of two lists' still has the cardinality $\aleph _0$ because the members of both lists come from the set of naturals and both lists can be combined to give back naturals. But this 'list of two lists' has order type $2\omega$.
We can keep doing this to produce new order types. We can take out every member at an even position in the second list and list those below our second list. This list of 'three lists' has an order type $3\omega$.
Continuing this, we'll have a list of infinite lists of cardinality still $\aleph _0$, but order type $\omega\cdot \omega=\omega^2$. We can still keep doing this to get order-types of $\omega^{\omega}$, $^{\omega}\omega$, etc. The first order-type which we cannot get by just re-ordering the naturals like this is $\omega _1$ and since we can't get there by re-ordering the naturals, then it has to have a higher cardinality than the naturals ($\aleph _1$).
This is my understanding of $\aleph _1$. Now my proof:
We first list out the countable list of real numbers of the form $2^{-n}$, where $n$ is a whole number ($1,\frac{1}{2},\frac{1}{4}.....$). This list has cardinality $\aleph _0$ and order type $\omega$. From each member of this list, we'll generate another list, so that we have a list of lists.
From each member $2^{-n}$, we generate the infinite list : $2^{-n}, \frac{2^{-n}+2^{-(n+1)}}{2}, \frac{\frac{2^{-n}+2^{-(n+1)}}{2} + 2^{-(n+1)}}{2},.....$, and so on.
Now we have an infinite list of infinite lists of cardinality $\aleph _0$ and order type $\omega ^2$. From each member of the sub-infinite lists, we can again use a process similar to above to generate an infinite list. For example, the first two members of the first sub-infinite list are $1$ and $\frac{3}{4}$. We can product the following infinite list from $1$:
$1, \frac{1+\frac{3}{4}}{2}, \frac{\frac{1+\frac{3}{4}}{2}+\frac{3}{4}}{2},.....$ and so on.
So continuing this, we can get higher order types $\omega^{3}, \omega^{\omega}, ^{\omega}\omega$....
The first order-type we can't get by continuing this process has to have the cardinality $\aleph _1$.
We see that we can get arbitrarily close to any real number in the range $(0,1)$ by using this process. Since we can get arbitrarily close to any of the members of $(0,1)$, so there can't be any set with cardinality in-between of the the cardinality of $(0,1)$ and the cardinality of naturals.
This is really just an elaboration of what's been said by others in the comments...
As commenters have noted, all you seem to have done here is put subsets of the rationals into various countable order-types. You haven't been very explicit about how to do this, but it's certainly true that the rationals can be put into any countable order type. The reason is trivial: the rationals are countable.
But then you jump the gun and say that since you imagine carrying the process to larger and larger countable order types will eventually produce a rational arbitrarily close to a given real, that means that there are no more than $\aleph_1$ reals.
I think you're imagining that the you are covering all of the reals by the various countable order types in this way, but notice there are no real numbers on your list at all, only rationals. (Also, there are a ton of different numbers on each level so it's unclear how you mean to claim there's a surjection from $\aleph_1$ here...)
Commenters have offered a bit of a reductio ad absurdum that I'll echo. There was no reason you needed the first stage in your process (order type $\omega$) to be $2^{-1}, 2^{-2},\ldots.$ The rationals are countable. You could have chosen an enumeration of the full list of rationals in type $\omega$. So, in one stage, you get arbitrarily close to every real number. So the reals are countable? (Or worse... they have cardinality 1??)
EDIT
I seem to have misunderstood your argument. The idea instead seems to be that somehow since you can produce a rational at each point of each countable ordinal, in some way that is vaguely "continuous" that the limit when you iterate this over the countable ordinals will be an ordering of type $\omega_1$ which has a real number at each point, and also contains every real number since you can somehow guarantee each real number is 'approached' (and thus "converged to").
There is a lot of handwaving here, to the point that this is almost too vague to form a clear picture of and refute. So instead I'll ask you to try to fill in the gaps... which real number is assigned to a given countable ordinal? Which real number is assigned to $\omega^2$ here? Can you actually show that for every real number, there is some countable ordinal that it is assigned to under the "limit"?
I don't anticipate you will have success here. As others have note, it has been proven some time ago that one cannot prove the CH from ZFC, much less from real analytic sort construction such as you're trying to do here.