What conditions allow a metric on an open ball to be completed to a closed ball? Given a metric space that is homeomorphic to an open $n$-ball, what conditions on the metric allow one to conclude that the completion is homeomorphic to a closed $n$-ball?
I know totally bounded is necessary, but it's hard to rule out weird cases like a closed ball with two points identified happening.
 A: As Mike Miller pointed out, coarse properties like being totally bounded will not suffice here; the nature of the completion depends on the fine structure of the metric near the boundary. In this spirit, I can offer   a mostly tautological statement: 
Fact. For a metric space $X$ that is homeomorphic to open ball $B^n$, the following are equivalent:


*

*The completion of $X$ is homeomorphic to closed ball $\overline{B}{}^n$

*$X$ is uniformly homeomorphic to $B^n$.


A uniform homeomorphism is a homeomorphism that is uniformly continuous with uniformly continuous inverse. Another way to express $(2)$ is that the original metric on $X$ is equivalent to a metric pushed over from $B$ in a uniform way: one being  $<\delta$ implies the other being $<\epsilon$, and vice versa.
It's clear that $1\implies 2$, by compactness of $\overline{B}{}^n$.
Conversely, a uniformly continuous map into a complete space extends continuously to the completion of the domain. Thus, a uniform homeomorphism $f:X\to B^n$ and its inverse $g:B^n\to X$ extend to continuous maps $F:\overline{X}\to \overline{B}{}^n$ and $G:\overline{B}{}^n\to\overline{X}$. By density, it follows that $F\circ G=\operatorname{id}_{\overline{B}{}^n}$ and $G\circ F=\operatorname{id}_{\overline{X}}$, as required.
