If $d_1$ and $d_2$ are metrics on the same set $X$ which satisfy the hypothesis that for any point $x \in X$ and $\epsilon >0$ there is a $\delta >0$ such that
$d_1(x,y)<\delta \implies d_2(x,y) < \epsilon$
and
$d_2(x,y)<\delta \implies d_1(x,y) < \epsilon$
then these metrics define the same open sets in $X$.
Here is my proof
Because we can choose $\delta \leq \epsilon$,
we can restate the hypothesis
$B_1(x;\delta) \subset B_2(x;\epsilon)$
and
$B_2(x;\delta) \subset B_1(x;\epsilon)$.
Let $U$ be a $d_1$ open set where $U=\cup_{x \in U} B_1(x;\epsilon)$ ,
so by assumption there is a $\delta$ such that $B_2(x;\delta) \subset B_1(x;\epsilon)$ .
This means each $x \in U$ contains a $d_2$ open ball, so by definition $U$ is also open in $d_2$.
By symmetric, we can prove $V=\cup B_2(x;\epsilon)$ open in $d_1$.
Best Answer
Here is a brief outline of the proof: it is sufficient to show that every $d_1$-open set is $d_2$-open set (why?)
Let $V$ be a $d_1$-open set. For each $x\in V$ we have a $d_1$-open ball $B_1(x,\varepsilon)\subseteq V$. By assumption there is $\delta$ such that $B_2(x,\delta)\subseteq V$ (how to get an inclusion?) so $V$ satisfies the definition of $d_2$-openness.