This question has some context, which I explain below.
Let $(X, d)$ be a metric space, and fix some $x_0$ in $X$. Define $BC(X) = \{ f : X \rightarrow \mathbb{R} : f~\text{is continuous and bounded}. \}.$ Equip $BC(X)$ with the supremum metric, that is, $\rho(f, g) = \sup_{x \in X} |f(x) - g(x)|.$ It's easy to see that $(BC(X), \rho)$ is a metric space. Now consider the map $\Phi : X \rightarrow BC(X)$ which sends $x$ to $\Phi_x$, where $\Phi_x(y) = d(x, y) - d(x_0, y).$ The map $\Phi$ is an isometry. I've got no problem up to this point.
Now, suppose $K \subseteq X$ is a compact set and $\{x_n\}_{n \geq 1}$ is a sequence in $X$ (not necessarily in $K$). My question is:
How can I show that there is a subsequence of $\{\Phi_{x_n}\}_{n \geq 1}$ that converges uniformly on $K$?
The first thing that popped in my head was, of course, Arzelà-Ascoli, but the family $\{\Phi_{x_n}\}_{n \geq 1}$ is not pointwise bounded, so I'm stuck. Any help is appreciated. Thanks.