How does $\text{sup}$ become a (co)limit in a subset of $\mathbb{R}$?
Does it actually become $\text{lim}$ or $\text{colim}$?
I.e. I wish, in the natural order $\leq$ in $\mathbb{R}$, that the thin
subcategory $X$ in $\mathbb{R}$ naturally satisfies $\sup X=\text{(co)lim} X$
to see that $\mathbb{R}$ is (co)complete in the categorical sense.
Any topological space $X$ can become a category, whose objects are the open sets in $X$, and the morphisms are $$\mathrm{Hom}(U,V) := \begin{cases} V\hookrightarrow U & \text{if } V\subseteq U, \\ \varnothing & \text{otherwise.} \end{cases}$$
I'm not sure if you should reverse the inclusions like that. Either way, if $X$ is a bounded subset of $\mathbf{R}$, try inducing the order topology on it, and then take the colimit of all of its open subsets. I think you should take $X$ to be closed as well, which isn't a problem since the supremum of $X$ equals that of $\overline{X}$.
Also, you might instead want to define a category whose objects are the closed subsets of $X$.