I tried proving that
Every bounded increasing sequence converges in $\mathbb{R}$.
implies that $\mathbb{R}$ has the least upper bound. Here, $\mathbb{R}$ is taken as an ordered field which contains the natural numbers.
I came to a conceptual problem in a particular step of my atempt, which I don't know if it is justified. I'll highlight it below:
Proof: Since you can prove the Archimedean property directly from MCT, it also follows easily the density of rationals.
(IDEA: We take an increasing sequence. We would want it to converge to the $\sup$, but it may not necessarily converge (indeed, it most likely will not). We proceed with stubbornness.)
Let $A$ be bounded by above, non-empty. Assume that there is no least upper bound of $A$.
Consider $\Omega$ (the first uncountable ordinal).
Define a function $f: \Omega \rightarrow \mathbb{Q}$ by transfinite recursion:
Take $f(0)$ to be any rational smaller than an element of $A$.
Given $f(a)$, take $f(a+1)$ to be a rational greater than $f(a)$ and smaller than an element of $A$.
$\color{red}{\text{Given a limit ordinal $\gamma$, suppose we have defined an increasing function on the ordinals}}$$\color{red}{\text{for all ordinals smaller than $\gamma$}}$. We have that there exists an increasing sequence of ordinals $\alpha_n$ smaller than $\gamma$ that converges to the limit ordinal (order topology). The associated $f(\alpha_n)$ is a bounded increasing sequence of real numbers, hence converge to a given real number $x$. Take a rational number which is greater than $x$, and smaller than an element of $A$ (this is guaranteed by the assumption that there is no least upper bound of $A$).
We have thus constructed an injection from $\Omega$ to $\mathbb{Q}$. Since $\mathbb{Q}$ is enumerable and $\Omega$ isn't, we have a contradiction.
My conceptual doubt comes from the red part... I am assuming not only that I have defined the function on the smaller ordinals, but also that the function satisfies something I want. Is this valid? Why?