Suppose we have an ordered set $(A,\le)$ with the property that, for any decreasing chain (sequence) $a_0 \ge a_1 \ge \dotsb$, there exists $n \in \mathbb{N}$ such that, for each $k \in \mathbb{N}$, we have $a_n=a_{n+k}$.
We have to prove that any non empty subset $B \subseteq A$ has a minimal element.
I'm new to order theory, so I can not figure out how to deal with this problem.
You need to use the Axiom of Choice, and in particular, its equivalent Zorn's Lemma, combined with the Recursion Theorem.
Assume that $B\ne \varnothing$ does not possess a minimal element. Picking an $a_0\in B$ you can define recursively the sequence $$ \text{$a_n$ is an element of $B$ strictly smaller than $a_{n-1}$}. $$ Such $a_n$ exists in $B$, since $a_{n-1}$ is, by assumption, not a minimal element, and this leads to a contradiction.