Lee's Intro to Topology, generating the same topology 
Suppose $M$ is a set and $d, d^\prime$ are two different metrics on
$M$. Prove that $d$, and $d'$ generate the same topology on $M$ if and
only if the following condition is satisfied: for every $x \in M$ and
every $r > 0$, there exists positive real numbers $r_1, r_2$ such that
$B_{r_1}^{(d^\prime)}(x) \subset B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x)
\subset B_r^{(d^\prime)}(x)$.

My attempt:
Suppose $d$ and $d'$ generate the same topology on $M$. The properties of a topology are that $\emptyset$ and $M$ are open subsets of $M$. An open subset means there exists an open ball $B_r^{(d)}(x) = \{y\in M : d(x,y) < r\}$ around each $x$.
The issue for me is choosing these $r_1, r_2$. I'm fairly new to topology and metric spaces so forgive me if this seems novice, but it is difficult to wrap my head around this and a lot of posts are using terms I'm unfamiliar with.
 A: I’ll do one direction and give you a chance to try your hand at the other direction once you’ve seen the kinds of reasoning involved. Suppose that for each $x\in M$ and $r>0$ there are $r_1,r_2>0$ such that $B_{r_1}^{(d')}(x)\subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x)\subseteq B_r^{(d')}(x)$; we’ll show that $d$ and $d'$ generate the same topology on $M$.
Let $\tau$ be the topology on $M$ generated by $d$ and $\tau'$ the topology generated by $d'$; we’ll show that $\tau=\tau'$ by showing that $\tau\subseteq\tau'$ and $\tau'\subseteq\tau$, so let $U\in\tau$. Because $\tau$ is generated by $d$, for each $x\in U$ there is an $r(x)>0$ such that $B_{r(x)}^{d}(x)\subseteq U$. By hypothesis for each $x\in U$ there is an $r_1(x)>0$ such that $B_{r_1(x)}^{(d')}(x)\subseteq B_{r(x)}^{(d)}(x)$. But then
$$U\subseteq\bigcup_{x\in U}B_{r_1(x)}^{(d')}(x)\subseteq\bigcup_{x\in U}B_{r(x)}^{(d)}(x)\subseteq U\;,$$
so $U=\bigcup_{x\in U}B_{r_1(x)}^{(d')}(x)$, which by definition is in $\tau'$. $U$ was an arbitrary member of $\tau$, so we’ve shown that $\tau\subseteq\tau'$. The proof that $\tau'\subseteq\tau$ is almost identical: start with an arbitrary $U\in\tau'$ and follow the same path, simply reversing the rôles of $d$ and $d'$ and this time using the fact that for each $x\in M$ and $r>0$ there is an $r_2>0$ such that $B_{r_2}^{(d)}(x)\subseteq B_r^{(d')}(x)$.
For the other direction you’ll assume that $\tau=\tau'$ and show that for each $x\in M$ and $r>0$ there are $r_1,r_2>0$ such that $B_{r_1}^{(d')}(x)\subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x)\subseteq B_r^{(d')}(x)$. Use the fact that by definition $B_r^{(d)}(x)\in\tau$, so by hypothesis $B_r^{(d)}(x)$ is also in $\tau'$. Similarly, $B_r^{(d')}(x)\in\tau'$ by definition, so by hypothesis $B_r^{(d')}(x)\in\tau$ as well.
