Let $X$ be a locally compact metric space, and suppose that the group of isometries of $X$ acts transitively. Show that $X$ is complete. (This is 2nd part of a problem. In first part I showed that for all $x\in X$ there exists $\epsilon >0$ such that $\bar B_\epsilon(x)$ is compact.)
Attempt:
Let $\{x_i\}$ be a Cauchy sequence in $X$. Let $x\in X$ and let $\epsilon$ be such that $\bar B_\epsilon(x)$ is compact. Our sequence is Cauchy so given we can choose $\epsilon/2 >\delta >0$, we can find an $N$ such that $d(x_i,x_j)< \delta$ for $i,j>N$. Let $z =x_k$, where $k>N$.
Since the group of isometries, $G$, acts transitively we have a $g\in G$ s.t $g(x)=z$, and $g(\bar B_\epsilon(x)=\bar B_\epsilon(z)$.Thus, $\{x_i\}, i>N$ is contained in $\bar B_\epsilon(z)$, which is compact and therefore sequentially compact, and so converges to $y \in X$.