Lipschitz Equivalent Metrics are Topologically Equivalent/Proof 2

Theorem
Let $M_1 = \left({A, d_1}\right)$ and $M_2 = \left({A, d_2}\right)$ be metric spaces on the same underlying set $A$.

Let $d_1$ and $d_2$ be Lipschitz equivalent.

Then $d_1$ and $d_2$ are topologically equivalent.

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

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

Let $x \in A$.

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

Let $B_{h \epsilon} \left({x; d_1}\right)$ be 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:
 * $\exists \epsilon \in \R_{>0}: B_\epsilon \left({x; d_1}\right) \subseteq U$

Thus:
 * $B_{\epsilon / k} \left({x; d_2}\right) \subseteq B_\epsilon \left({x; d_1}\right) \subseteq U$

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.