Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2

Theorem
Let $R$ be a division ring.

Let $\norm{\,\cdot\,}_1: R \to \R_{\ge 0}$ and $\norm{\,\cdot\,}_2: R \to \R_{\ge 0}$ be norms on $R$.

Let $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ satisfy:
 * $\forall y \in R:\norm{y}_1 \lt 1 \iff \norm{y}_2 \lt 1$

Let $x, x_0 \in R \setminus 0_R$ be such that $\norm {x}_1 = \norm {x_0}_1$.

Then:
 * $\norm {x}_2 = \norm {x_0}_2$

Proof
Since $x \neq 0_R$ then $x$ has a product inverse $x^{-1}$.

Similarly $x_0$ has a product inverse $x_0^{-1}$.

Then:

and similarly:

Aiming for a contradiction, suppose that $\norm {x}_2 \neq \norm {x_0}_2$.

By Real Numbers form Totally Ordered Field either:
 * $\norm {x}_2 \lt \norm {x_0}_2$

or:
 * $\norm {x_0}_2 \lt \norm {x}_2$.

Hence either:

or similarly:

This contradicts the premise:
 * $\forall y \in R:\norm{y}_1 \lt 1 \iff \norm{y}_2 \lt 1$

So $\norm {x}_2 = \norm {x_0}_2$.