Is an expanding map on a compact metric space continuous?

I got inspired by this question Existence of convergent subsequence to think about the following problem: Suppose you have a compact metric space $(X,d)$ and an expanding map $T:X\rightarrow X$, i.e. $d(Tx,Ty)\geq d(x,y)$ for every $x,y\in X$. Is the map $T$ then continuous?

Without assuming continuity we already know that $T^n(X)$ must be dense in $X$ for any $n\geq 1$, since given $z\in X$ the orbit of $z$ must accumulate upon $z$. But showing continuity has escaped me so far. If there are counter-examples using the Axiom of Choice that also interests me.

Such a map $T$ must in fact be an isometry. Indeed, given $x,y\in X$, let $x_n=T^n(x)$ and $y_n=T^n(y)$. By compactness, we can find $n_k$ such that both $(x_{n_k})$ and $(y_{n_k})$ converge, and by passing to a further subsequence we may assume $n_{k+1}-n_k$ is increasing. Setting $m_k=n_{k+1}-n_k$, we then find as in your answer to the linked question that $(x_{m_k})$ converges to $x$ and $(y_{m_k})$ converges to $y$. In particular, $d(x_{m_k},y_{m_k})$ must converge to $d(x,y)$. But as long as $m_k>0$, $d(x_{m_k},y_{m_k})\geq d(T(x),T(y))$. It follows that we must have $d(T(x),T(y))\leq d(x,y)$.

(By compactness of $X$ and your observation that $T(X)$ is dense, it follows that $T$ is also surjective.)