Question About equivalent metrics Suppose $\rho$ and $\sigma$ are two equivalent metrics on a set $X$, $y \in X$ and $\epsilon$ and $\delta$ are positive numbers if $ \overline {B_{\rho}(y,\epsilon)}$ is compact and is contained in $B_{\sigma}(y,\delta)$ then show that there exist an r > 0 such that $\overline {B_{\rho}(y,\epsilon + r)} \subseteq B_{\sigma}(y,\delta - r)$.
 A: You can’t use the boundary of $B_\sigma(y,\delta)$ for anything, because it may be empty: $B_\sigma(y,\delta)$ may be clopen. But there’s a bigger problem: the theorem is false.
Let $X=\mathbb{Z}\times [0,1]$. Let $\sigma$ be the restriction to $X$ of the usual Euclidean metric in the plane, and let $\rho$ be the metric defined as follows:
$$\rho(\langle m,x\rangle,\langle n,y\rangle)=\begin{cases}
|x-y|,&m=n\\
1,&m\ne n\;.
\end{cases}$$
Let $p=\langle 0,0\rangle$; then $B_\rho(p,1)=\{0\}\times [0,1)$, and $B=\overline{B_\rho(p,1)}=\{0\}\times[0,1]$, which is compact. 
Clearly $B\subseteq B_\sigma(p,2)=\{-1,0,1\}\times[0,1]$, $B_\sigma(p,2-r)\subseteq B_\sigma(p,2)$ for any $r\in (0,2]$. But $B_\rho(p,1+r)=X$ for any $r>0$, so there is no $r>0$ such that $\overline{B_\rho(p,1+r)}\subseteq B_\sigma(p,2-r)$.
A: As Brian has pointed out, the result is false. Here is a perhaps slightly simpler counterexample, where even $\rho=\sigma$.
Take $X= \{0\}\cup \{1+1/n:n\in\mathbb{N}\}$, and let $\rho=\sigma$ be the Euclidean metric on $X$. Since both metrics are the same, I will drop the subscripts.
Now $\overline{B(0,1)}=\{0\}$ is compact and contained in $B(0,1)$. However, $\overline{B(0,1+\varepsilon)}$ is not contained in $B(0,1)$ for any $\varepsilon>0$.
(This example was written with the assumption that the OP used $\overline{B(x,r)}$ to denote the closed ball of radius $r$ around $x$, rather than the closure of the open ball. Otherwise, an even easier example is given by $X=\{0,1\}$.)
