Showing that the following two metric are equivalent. Let $(X,d)$ be a compact metric space, and $f \colon (X,d)\to \mathbb{R}$ is a continuous function such that if $x,y \in X$ and $x \neq y$ than $f(x) \neq f(y)$.
Let $t \colon X  \times X \to \mathbb{R}$ where $t(x,y) = |f(x)-f(y)|$.
I want to prove that $t$ is a metric and that $t$ and $d$ are equivalent. 
Proving that $t$ is a metric is easy. In order prove that these two are equivalent, I am thinking about showing that every subset of $X$ is $d$-closed if and only if it is $t$-closed, considering that every singleton set $\{x\}$ is closed in both $(X,d)$ and $(X,t)$. Is this a correct way of thinking ?
 A: Your approach won't bring you very far. Singletons are closed in every metric space, so this doesn't help you much. Also, the compactness of $X$ is crucial for the proof, while your approach doesn't use it at all.
Try showing that every $t$-open set is also $d$-open and vice versa.
It is easy to show that every $t$-open set is $d$-open. This follows directly from the $\epsilon$-$\delta$-characterization of continuity and doesn't even need the compactness of $X$.
For the other direction, consider the inverse function $f^{-1}: f(X) \to X$. Since $(X, d)$ is compact, this function is continuous*. Now let $U \subset X$ be $d$-open and $x \in U$. Then there exists a $\delta > 0$ so that $U_\epsilon(x) \subset U$. Now choose an $\epsilon > 0$ so that $f^{-1}(U_\epsilon(f(x))) \subset U_\delta(f^{-1}(f(x))) = U_\delta(x)$ [this is the continuity of $f^{-1}$ at the point $f(x)$]. But this means $t(x, y) < \epsilon \implies  d(x, y) < \delta$ (!), i.e. the $\epsilon$-neighborhood of $x$ with respect to $t$ is a subset of $U_\delta(x) \subset U$. Therefore we've shown that $U$ is $t$-open.
(*) To see this, observe that a function is continuous iff the preimage of every closed set is closed. If $M$ is closed, then it is also compact [since $X$ is compact] and therefore $(f^{-1})^{-1}(M) = f(M)$ is compact. Since $\mathbb{R}$ is Hausdorff, this implies the closedness of $f(M)$.
