How to show whether two metrics generate the same topology

I’ll write $d_E$ for the usual Euclidean metric on $\Bbb R$. It suffices to show two things:

1. For each $x\in\Bbb R$ and each $\epsilon>0$ there is a $\delta>0$ such that $B_d(x,\delta)\subseteq B_{d_E}(x,\epsilon)$.
2. For each $x\in\Bbb R$ and each $\epsilon>0$ there is a $\delta>0$ such that $B_{d_E}(x,\delta)\subseteq B_d(x,\epsilon)$.

If you’ve learned about bases for topologies, you may recognize this as showing that the the open $d$-balls and the open $d_E$-balls are bases for the same topology on $\Bbb R$.

The reason that this works is simple. Suppose that $U$ is open in the Euclidean topology. Then for each $x\in U$ there is an $\epsilon_x>0$ such that $B_{d_E}(x,\epsilon)\subseteq U$. If you’ve shown (1) above, you know that for each $x\in U$ there is a $\delta_x>0$ such that $B_d(x,\delta_x)\subseteq B_{d_E}(x,\epsilon_x)\subseteq U$, and therefore $U$ is open in the topology generated by $d$ as well.

The argument in the other direction is exactly the same, except that you use (2) instead of (1).

For now I’ll leave it to you to try to prove (1) and (2).