There is an exercise on page 14 in the Tomas Jech's "Set Theory":
Every nonempty $X \subseteq \mathbb{N}$ has an $\in$-minimal element. [Pick $n \in X$ and look at $X \cap n$.]
and there is a good solution written by the user egreg: https://math.stackexchange.com/a/1526646/251394 . (We deal with the intersection of all elements of the set.)
This proof has a lot of inductions. For example, we need to prove trichotomy. And induction itself is not yet proved. That is strange because other theorems nearby (1.1-1.6 and 1.8-1.9) are very easy, relatively to 1.7.
Could you please suggest a shorter proof?
p.s. We also can use the regularity axiom, but it is not fair.
p.p.s. I think, it is sufficient to show that $\bigcap X \in X$. (Then proof by contradiction: assume $s\in X \cap \bigcap X$, so $s\in X$ and $\forall m\in X.s\in m$, which means that $s\in s$, so there is a contradiction by ex.1.5)
By exercise 1.6 you have that for every element $n$ of $\mathbb{N}$, if $z\subseteq n$ is not empty, then has a $\in$-minimal element. So, you have two cases
$X\cap n=\emptyset$: In this case $n$ is a $\in$-minimal element of $X$
$X\cap n\neq\emptyset$: As $X\cap n\subseteq n$ then exists a $\in$-minimal element $m\in X\cap n$. So, $m$ is $\in$-minimal in $X$ too. (Edited: in the last sentence of the second case we use ex.1.4, which states that every natural number is transitive.)