How to show that every nonempty $X \subset \mathbb N$ has an $\in$-minimal element I am trying to prove the following:

Every nonempty $X \subset \mathbb N$ has an $\in$-minimal element.

Proof: Take some $n \in X$. Then $\min \{ n \cap X \} = \min \{ X \}$.
Is this ok or should it be more detailed? 
 A: As to whether your answer is OK in the sense of being correct, I'd say yes.
As to whether it's detailed enough, I'd say: That depends.  There are a lot of seemingly "easy" questions like these in the beginning of set theory and the most important point is not to find some kind of "acceptable" answer but rather to watch out for what you already know (by axioms or things you've already proved) as opposed to what you only think you know.  I'd therefore recommend that in the beginning you are as detailed as possible justifying each step in your proof.  And for others to help you you need to tell them exactly this: What you already know and are therefore allowed to use (like from previous exercises).  Once you get the hang of it, you can get a bit more loose.
Here for example for $\min\{n \cap X\}$ to make sense you already need to know that $x \in \mathbb N$ implies $x \subseteq \mathbb N$.  And then you need to know that $n$ (viewed as a set of natural numbers) has an $\in$-minimal element.  For someone trained in set theory, these are all obvious facts, but they won't know in which order you've been taught such things, i.e. whether you are allowed to use these facts or not.  That makes it difficult to help you.
