Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 2

Proof
By definition of Lipschitz equivalence:
 * $\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$

for some $h, k \in \R_{>0}$.

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_{h \epsilon} } {x; d_1}$ denote the open $h \epsilon$-ball with respect to $d_1$ of $x \in A$.

Then:

Similarly:

Now suppose $U \subseteq A$ is $d_1$-open.

Let $x \in U$.

Then:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$

Thus:
 * $\map {B_{\epsilon / k} } {x; d_2} \subseteq \map {B_\epsilon} {x; d_1} \subseteq U$

and so $U$ is $d_2$-open.

Mutatis mutandis, if $U \subseteq A$ is $d_2$-open, it follows that $U$ is $d_1$-open.