Showing a homeomorphism has dense orbit for every x in a compact metric space X, if $d(x,y)=d(T(x),T(y))$ I am self studying entropy/topological dynamics, and came across this problem.
Note: Two definitions: 
1)We say T is topologically transitive if there exists an $x\in X $ s.t.$ Orb(x):= \{T^k(x): k\in \mathbb{Z}\}$ is dense in X, i.e. if its closure is $X$.
2)We say T is minimal if every x in X has dense orbit.
The problem: 
Let X be a compact metric space with metric d and let $T : X → X$ be a topologically transitive homeomorphism. Show that if T is
an isometry (i.e. $d(T(x), T(y)) = d(x, y)$, for all $x, y ∈ X$) then T is minimal.
I have tried the following. Choose an arbitrary y in X, let $\epsilon>0$ arbitrary and then show that for arbitrary z in X, there exists a $k\in \mathbb{Z}$ s.t. $d(T^k(y),z)<\epsilon$ by using triangle inequality and the isometry but I just can't get anywhere. Could anyone help? Thanks in advance!
 A: Assume $T$ is a transitive isometry.

The goal is to show that for all $x \in X$, the orbit of $x$ is dense in $X$.

Fix $x,y \in X$.

We want to show $y$ is in the closure of the orbit of $x$.

Equivalently, we want to show that for all $\epsilon>0$, there exists $p \in \mathbb{Z}$ such that $d(T^p(x),y) < \epsilon$.

Fix $\epsilon>0$.

Since $T$ is transitive, there exists $w \in X$ such that the orbit of $w$ is dense in $X$.

Hence there exist $m,n \in \mathbb{Z}$ such that
$$d(T^m(w),x) < \frac{\epsilon}{2}$$
$$d(T^n(w),y) < \frac{\epsilon}{2}$$
Let $p=n-m$.
\begin{align*}
\text{Then}\;\;d(T^p(x),y) &= d(T^{n-m}(x),y)\\[4pt]
&=d(T^{-m}(x),T^{-n}(y))&&\text{[since $T$ is an isometry]}\\[4pt]
&\le d(T^{-m}(x),w)+d(w,T^{-n}(y))\\[4pt]
&=d(w,T^{-m}(x))+d(w,T^{-n}(y))\\[4pt]
&=d(T^m(w),x)+d(T^n(w),y)&&\text{[since $T$ is an isometry]}\\[4pt]
&<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\[4pt]
&=\epsilon\\[4pt]
\end{align*}
Hence $y$ is in the closure of the orbit of $x$, as required, which thus completes the proof.

Note: I never used compactness.
A: Let $x\in X$ have dense orbit, $y,z\in X$, and $\varepsilon>0$.
Consider the sequence $(T^k(y))_{k=1}^\infty$. Due to sequential compactness, we can find natural numbers $n_1<n_2$ such that $d(T^{n_1}(y),T^{n_2}(y))<\varepsilon/3$. Using the fact that $T$ is an isometry, we deduce $d(T^k(y),y)<\varepsilon/3$ for some natural number $k$.
Since $x$ has dense orbit, there exists integers $m$ and $n$ satisfying
$d(T^m(x),y)<\varepsilon/3$ and $d(T^n(x),z)<\varepsilon/3$.
Then
$$
d(T^{k+n-m}(y),z)\leq
d(T^{k+n-m}(y),T^{n-m}(y))+d(T^{n-m}(y), T^{m+n-m}(x)) +d(T^n(x),z) \\
= d(T^k(y),y)+d(y, T^m(x)) +d(T^n(x),z) < \varepsilon.
$$
Therefore $T$ is minimal.
