I am recalling the definition of Harnack distance. Let $D$ be a subdomain of $\mathbb{C}$. Then it is true that for every pair $z,w\in D$, there exists a number $\tau>0$, so that $$\tau^{-1}\cdot h(z)\leq h(w)\leq\tau \cdot h(z)\qquad\forall h\text{ positive harmonic on }D.$$ We call Harnack distance the smallest number $\tau$ with the aforementioned property and denote it as $\tau_{D}(z,w)$.
By a known Theorem, Theorem 1.3.8 Ransford Potential Theory, it is true that $$\log\tau_{D}\text{ is a semimetric, which is continuous from }D\times D\rightarrow \mathbb{R}$$ where continuous here means with respect to the respected Euclidean Topology.
Let us now assume that $D$ is a bounded subdomain, in particular assume that its (Euclidean) diameter equals $\delta>0$. Then we can prove that $$\tau_{D}(z,w)\geq \frac{\delta+|z-w|}{\delta-|z-w|}\qquad (z,w\in D).$$ It is clear that this condition makes $\log\tau_D$ a proper metric on $D$. However, it is stated also that now the Topology induced from $\log\tau$ is the same as the Euclidean one. I cannot prove that statement, specifically I cannot prove it in its full context. It is clear to me that since $\log\tau_{D}$ is continuous, that if for a pair $(z,w)\in D\times D$, we have that $$\log\tau_{D}(z,w)<\varepsilon\qquad\text{ for some choice of }\varepsilon>0$$ then continuity implies we can find a $\delta>0$, such that $w\in B_e(z,\delta)\iff |z-w|<\delta$. In particular, we have that $$\forall \varepsilon >0\quad\exists \delta>0,\Rightarrow \forall z\in D\colon B_{e}(z,\delta)\subset B_{\log\tau_{D}}(z,\varepsilon)$$ and we know from the Theory of Metric spaces that this condition implies that the Euclidean Topology is a subcollection of the Topology induced from Harnack distance.
Does anyone have any ideas or hints for the other inclusion? Also if I am wrong in the part which I showed now, I am happy to hear any comments that corrects this question.
Best regards,