When do isometric metrics induce the same topologies?

Question

It is well-known that strongly equivalent metrics induce equivalent topologies (i.e. the topologies are homeomorphic). However, I was wondering about the question of same topologies and not just equivalent. I think I have proven the following result in this direction:

Let $d$ and $d'$ be metrics on a set $X$ with $f\colon (X, d)\to (X, d')$ being an isometry. Then the following hold:

1. $\mathcal T_d\subseteq \mathcal T_{d'}$ $\iff$ $f\colon (X, d)\to (X, d)$ is continuous.
2. $\mathcal T_d\supseteq\mathcal T_{d'}$ $\iff$ $f^{-1}\colon (X, d)\to (X, d)$ is continuous.

How will this help? We can hence take a discontinuous bijection $f$ on $(X, d)$ and then even though $(X, d)$ and $(X, d')$ will be strongly equivalent (in fact, isometric), their topologies will not be the same since $\mathcal T_d\nsubseteq\mathcal T_{d'}$.

My attempt at the proof:

Note that since $f$ is an isometry, $f\colon (X, d)\to (X, d')$ and $f^{-1}\colon (X, d')\to (X, d)$ are both continuous.

1. It suffices to show that the identity function $\iota\colon (X, d')\to (X, d)$ is continuous $\iff$ $f\colon (X, d)\to (X, d)$ is continuous. "$\Rightarrow$" follows since the composition of $f\colon (X, d)\to (X, d')$ and $\iota\colon (X, d')\to (X, d)$ will be continuous. "$\Leftarrow$" follows since the composition of $f^{-1}\colon (X, d')\to (X, d)$ and $f\colon (X, d)\to (X, d)$ will be continuous.

2. It suffices to show that $\iota\colon (X, d)\to (X, d')$ is continuous $\iff$ $f^{-1}\colon (X, d)\to (X, d)$ is continuous. "$\Rightarrow$" follows since the composition of $\iota\colon (X, d)\to (X, d')$ and $f^{-1}\colon (X, d')\to (X, d)$ will be continuous. "$\Leftarrow$" follows since the composition of $f^{-1}\colon (X, d)\to (X, d)$ and $f\colon (X, d)\to (X, d')$ will be continuous.

Question: Does the above proof look okay?

Answer 1

$f:(X, d) \to (X, d') $ isometry i.e $$d'(f(x), f(y)) =d(x, y) $$

Claim 1 : An ismoetry is necessarily injective.

Proof : Suppose $f$ is an isometry.

Then $f(x) =f(y) $ implies $d(x,y)=d'(f(x), f(y)) =0$

Hence $x=y$

Calim 2 :An onto isometry is a homeomorphism.

Proof : Suppose $f$ is an onto ismoetry then $f$ is bijective and $f^{-1}$ is also an ismoetry. So it is enough to show $f$ is continuous and which is trivial .

Hence $(X, d) $ and $(X, d') $ are homeomorphic i.e $\tau(d) =\tau(d') $.

Then $id : X\to X $ is clearly both $d-d'$ and $d'-d$ continuous.