For $n\geq 1$ let $K_n$ be the partially ordered set of compact subsets of $\mathbb{R}^n$ ordered by inclusion. Let $B_n$ be the totally ordered set of closed balls in $\mathbb{R}^n$ centered at the origin with radius a positive integer (ordered by inclusion). Consider $K_n$ as a thin category $\underline{K_n}$ and $B_n$ as a thin category $\underline{B_n}$. Is the inclusion functor $I\colon \underline{B_n}\rightarrow \underline{K_n}$ a final functor? Is there a (high level) abstract-nonsense argument for this?
If $I$ was a right adjoint, this would follow directly (see example 5.2 in the nLab article). I think $I$ is a right adjoint, but do not know what the left adjoint is explicitly. $I$ preserves limits (set union is the limit in both categories). $B_n$ is locally small and for $S\in \underline{K_n}$ the set $\{B_r(0)\}_{r>s}$ (where $s=\operatorname{max}_{x\in S}\vert \vert x \vert \vert$) with the inclusion of $S \subseteq B_r(0)$ is a solution set. By the general adjoint functor theorem $I$ has a left adjoint — namely?
$I$ is a right adjoint; its left adjoint sends a compact subset of $\mathbb{R}^n$ to the smallest ball with integer radius containing it. Compare to the example of the inclusion $\mathbb{Z} \to \mathbb{R}$ which is both a left and a right adjoint, with its left adjoint given by ceiling and its right adjoint given by floor.