A set $A$ is minimal if it is nonempty, closed, invariant (i.e. $f(A) \subseteq A$) and it does not contain any proper nonempty, closed, invariant subset.
Let $X$ be a compact metric space and $f:X \to X$ a continuous map. Prove that $X$ contains some minimal set without using axiom of choice (or, equivalently, Zorn's lemma).
My attempt: if $X$ contains a periodic point, we are done, so assume $X$ does not contain any periodic point. Let $x_0 \in X$ and consider the set $\omega_f(x_0)$ (set of limit points of $Orb^+_f(x_0)$). Fix some $x_1 \in \omega_f(x_0)$ and consider the set $\omega_f(x_1)$. Fix some $x_2 \in \omega_f(x_1)$ and so on. Let $A = \bigcap_{n = 0}^{\infty} \omega_f(x_n)$. $A$ is an intersection of countably many closed sets, hence it is closed. Since $\omega$-limit sets are invariant and $\omega_f(x_{i + 1}) \subseteq \omega_f(x_i)$, it is invariant. Since $X$ does not contain any periodic point, $A$ is infinite.
I struggle to prove that it does not contain any nonempty, closed, invariant subset. And even if I could do that, I feel like I have already used the axiom of choice in picking $x_n$'s.