Limit construction process that avoids constructing a decimal like $x.99999999\ldots$ 
Each monotonically increasing sequence of real numbers that is bounded from above has a limit.

Below is a constructive proof of the statement above.
Let $(a_n)_{n≥1}$ be a monotonically increasing, bounded sequence where $a_n = A^{(n)}.\alpha_1^{(n)}\alpha_2^{(n)}\alpha_3^{(n)}\ldots$ with $A^{(n)} \in \mathbb N \cup \{0\}$ and $\alpha_j^{(n)} \in \{0, 1, 2, 3, \ldots, 9\}.$ By assumption, there's $T \ge 0$ s.t. $A^{(n)} < a_n \le T$ for all $n$. Now, $(A^{(n)})_{n\ge 1}$ is monotonically increasing, but it cannot fly off to infinity as it is bounded by $T$ so there must be some $n_0 \in \mathbb N$ s.t. $A^{(n)} = A$ for all $n \ge n_0.$ Similarly, there's some $n_1 \in \mathbb N$ s.t. $\alpha_1^{(n)} = \alpha_1$ for all $n \ge n_1$.
Using the algorithm above, we construct a number $a = A.\alpha_1\alpha_2\alpha_3\ldots$ The last step is to show $a = \lim_{n \to \infty}a_n$.
My question:
Consider $(\alpha_1^{(n)})$. One possible $(\alpha_1^{(n)})$ is $1, 1, 1, 1,\ldots, 1, 2, 3, 6, 8, 8, 8, \ldots, 9, \ldots$. Since $(\alpha_1^{(n)})$ is monotonically increasing, if/when it reaches, say, $8$, it never turns back and takes a value $< 8$, so it has to keep growing and eventually must reach $9$ and stay there.
So, how does this construction avoid the limit $A.999999\ldots?$
 A: This is a good intuitive approach, but it requires you to prove a lot of intuition about decimal notation.
There is no reason to avoid $A.99999\cdots,$ because that is a real number, too. If you need to avoid that, you’d also need to avoid $A.399999\dots$ or any number that is “eventually” all $9$s.
The real problem is that this proof assumes so much.
In our youth, we are taught about real numbers in terms of their base $10$ notation, and we think of real numbers, intuitively, as base $10.$
But in real analysis, we have to prove things about base $10.$
So, for example, you assume that if $a_0.a_1a_2\cdots<b_0.b_1b_2\cdots$ then for some $k,$ $a_k<b_k$ and $a_i=b_i$ for all $i<k.$
That seems intuitive, but it requires proof. (And we know this doesn’t go the other way, because: $0.999\cdots=1.000\cdots.$)
Even the notion, used in your proof, that every real number can be written as a decimal number, requires proof.
You also use the notion that every decimal expansion converges. That is practically a circular argument, because the usual real analysis proof that decimals converge uses that bounded monotonic increasing sequences converge. You might be able to prove the special case that all decimal numbers converge without the general case, but that seems hard to me.
So, this is a nice intuitive approach to understanding the theorem in base $10,$ but as a real analysis proof, it has a lot of holes. It might be able to be filled in, by filling in all those intuitions about base $10$ representations of real numbers, but it takes some work to fill them in.
What you find, if you try this, is that these are very messy proofs. The real analysis proof, using the supremum of the sequence, is much simpler.
At heart, you should be learning in calculus/real analysis that all those intuitions about real numbers have proofs. Base $10,$ or any base, is just one way of thinking about real numbers. But the base $10$ representation is not the heart of what the real numbers are, despite what we learned in grade school and high school.

One particular way that you assume decimal numbers are real numbers is in the assumption that a real numbers can constructively be turned into a decimal representations. In fact, constructively, you can’t always tell if a real number $a$ even satisfies $a<1,$ or $a\geq 1.$
But even if you are given a sequence of real numbers in the form of decimal numbers, you can’t constructively determine $n_0,n_1,\dots,$ so you can’t construct your limit.
So this is not a “constructive proof.” As Mark Saving’s answer notes, there can’t be a constructive proof.
A: 
Below is a constructive proof of the statement above.

Given that the statement above does not have a constructive proof, I highly doubt that.
For suppose that every monotonically increasing bounded sequence has a limit.
Given a Turing machine $M$ and an input $i$, define a sequence $\{a_n\}_{n \in \mathbb{N}}$ by
$$a_n = \begin{cases}
  0 & \text{$M$ does not halt on $i$ in $n$ or fewer steps} \\
  1 & otherwise \end{cases}$$
Now $a$ is a monotonically increasing, bounded sequence. Take its limit $L$. Note that if $M$ halts on input $i$, then $L = 1$.
Take some $N$ such that $|a_n - L| < 1$.
If $a_n = 0$, then it cannot be the case that $L = 1$, since $|0 - 1| = 1$ is not less than $1$. Therefore, $L \neq 1$, and thus $M$ does not halt on input $i$.
If $a_n = 1$, then $M$ does halt on input $i$.
Therefore, $M$ either halts on input $i$ or it does not.
However, it is impossible to prove constructively that for all Turing machines $M$ and all inputs $i$, $M$ either halts on $i$ or does not halt (unless the constructive metatheory used is flat-out inconsistent). Therefore, it is impossible to constructively prove that all monotonically increasing bounded sequences have a limit. $\square$
The first way that your argument is nonconstructive is that there is no constructive proof that every real number has a decimal representation. The second, more serious way is in your argument that there must be some $n$ such that $A^{(n)} = A$, which is also nonconstructive.
However, classically, your proof is acceptable. Recall that it's perfectly legitimate for a decimal representation to contain an endless string of nines, and that $0.\bar{9} = 1$. The rational numbers with finite decimal representations also have an infinite one involving an endless string of nines.
A much, much easier approach is to simply use Dedekind completeness to take the supremum of the sequence.
