I was trying to understand this article about the existence of complete Riemannian metrics by Nomizu Ozeki, see The Existence of Complete Riemannian Metrics, Proceedings of the American Mathematical Society, Vol. 12, No. 6, pp. 889-891, 1961.
I understand the general idea of building this function that increases indefinitely as you approach the missing points. So in the article, they define this function $r:M \rightarrow \mathbb{R}$, which is the supremum of positive numbers such that the ball $\mathbb{B}(x,r) = \{y\in M: d(x,y)<r\}$ is relatively compact, where the distance function is induced by the original Riemann metric on $M$: $$ d(x,y) = \inf \{\int\sqrt{\langle\frac{dc}{dt},\frac{dc}{dt}\rangle}dt: c \text{ piecewise differentiable curve joining } x \text{ and } y \}. $$
We may assume $r$ is always finite; otherwise, the manifold is already complete, as I understand it.
From here, we choose a differentiable function satisfying $w(x)>\frac{1}{r(x)}$ for each $x\in M$, via partitions of unity. Then the conformal metric is defined as $g'(x) = w(x)^2g(x)$ at each $x$. From here, everything seems to go smoothly. My difficulty is showing that the function $r:M \rightarrow \mathbb{R}$ is Lipschitz with respect to the metric $d$. Specifically, I wish to show that $$ |r(x)-r(y)|\leq d(x,y) $$ for all $x,y\in M$. Any ideas will be appreciated.
The key observation is that subsets of relatively compact are relatively compact: if $X \subseteq Y$ and $Y$ is relatively compact, then $\overline{X} \subseteq \overline{Y}$ is a closed subset of a compact set, hence is compact. So, if $B(x, d(x, y)+a)$ is relatively compact, then $B(y, a)$ is as well. Suppose for a contradiction that $r(x) > r(y) + d(x, y)$. Then for some $\epsilon> 0$ we have $a = r(x)- \epsilon > r(y) + d(x, y)$, hence $a - d(x, y) > r(y)$. By definition of $r$ we see that $B(x, a)$ is relatively compact whereas $B(y, a-d(x, y))$ is not, contradiction. The other direction follows by symmetry.