$f:X\to X$ is one-one and continuous on a compact space. Is $f$ surjective?

Question

Let $(X,\mathcal T)$ be a compact Hausdorff topological space and $f:X\to X$ be one-to-one and continuous. Is $f$ surjective?

Answer 1

No. Consider $X=[0,1]$ and $f(x)=\dfrac x2$. This is clearly continuous, one-to-one and $f^{n+1}(X)\subseteq f^n(X)$, but it is certainly not surjective.

Answer 2

If $X$ is metric (and compact), and if $d(f(x), f(y)) \geq d(x,y)$, then $f$ is (of course) injective, and you can show that $f$ is also surjective.

Hint :

Show that x is the limit of a subsequence of $(f^n(x))_{n \in \mathbb{N}}$. Which is, I suppose, the same idea of your proof when $X$ is a topological cancellative semigroup.