My approach (in a different way): We write $\mathbb{R}=\displaystyle\bigcup_{n \in \mathbb{Z}} [n-1,n]$ and let us call $[n-1,n]=I_n$.
Suppose $S \subset \mathbb{R}$ is uncountable and is well-ordered. Then, at least one of $I_n$'s, say $I_k$ will contain uncountably many elements of $S$.
$A=I_k \cap S$ is bounded and uncountable $\implies$ There exists at least one strictly monotone decreasing sequence $\{x_n\}$ in $A$ which converges to $p$ (say) [Justification of the claim below]. $\{x_n\} \subset S$ has no least element, since if $m= \displaystyle \min_{n\in \mathbb{N}} \{x_n\}=x_k$, we have $x_{k+1}<x_k=m$, a contradiction.
Thus, there is no such $S \subset \mathbb{R}$.
Justification:
https://math.stackexchange.com/a/3569035/389992
Is the procedure correct?
Kindly VERIFY