Let $(X,d)$ be a metric space. Then one knows that the initial topology on $X$ generated by the space of bounded continuous functions $C_b(X)$ with values in $\mathbb{R}$ coincides with the metric topology.
Now let $C_b^d(X)\subseteq C_b(X)$ denote the subspace of functions $f$ such that
$$\lim_{x\rightarrow\infty}\text{diam}(f(B(x,r)) = 0$$
for all $r>0$.
I have seen it claimed that if the metric $d$ is proper, then $C_b^d(X)$ also generates the metric topology. Why is this true?
Thoughts: I understand that in a proper metric space, closed and bounded sets (in particular closed balls of finite radius) are compact, so one might try and show that every compact subset of $X$ is a pre-image under some $f$ of a closed set in $\mathbb{R}$. But I'm not sure if this is the right approach, or how to use the properness condition.
Thanks!
A subset $F\subseteq C_b(X)$ generates the topology on $X$ if and only if $F$ separates points from closed sets, in the sense that given a closed subset $Y\subseteq X$ and a point $x\in X\backslash Y$, one can find a function $f\in F$ such that $f(x)\notin\overline{f(Y)}\subseteq\mathbb{R}$.
Thus the problem is to find an $f$ that maps an arbitrary point outside of the closure of the image of the arbitrary closed set $Y$. Naively, one could hope to map $x$ to a larger value than $\overline{f(Y)}$. One possible construction that will control the diameter of $\overline{f(Y)}$ is the following definition of $f$:
$$f: y\mapsto\frac{1}{1+d(y,x)},\qquad \forall y\in X.$$
Continuity of $f$ is ensured by continuity of $d$. Properness of the metric means that the compact sets are the closed and bounded sets, and hence $f$ is in $C_b^d(X)$ because it has vanishing variation with respect to radii of balls. The fact that $f(x)\notin \overline{f(Y)}$ can be shown as follows: take a closed annulus $A$ centred at $x$ such that $$\sup_{x\in A}\{f(x)\}\geq\sup_{x\in Y}\{f(x)\},$$ which is possible because the closed sets $x$ and $Y$ are separated by some distance $\delta > 0$. Then by properness $A$ is compact and hence $f(A)$ is compact in $\mathbb{R}$, so that it is separated by some distance $\eta > 0$ from $f(x)$.