$\mathbb{N} \in \mathbb{N}$? My professor proved this:
$\forall n \in \mathbb{N}: n \not\in n$
I proved that: $ n \bigcup \{n\} \not\subset n$
pf.
$Suppose,  n \bigcup \{n\} \subset n$
$\implies  (n \bigcup \{n\})\setminus n  \subset n\setminus n$
$\implies  n\setminus n \bigcup \{n\}\setminus n  \subset n\setminus n$
$\implies \emptyset \bigcup \{n\}\setminus n  \subset \emptyset$
$\implies\{n\}\setminus n  \subset \emptyset$
$Suppose, \{n\}\setminus n = \emptyset$
$\implies n \in n$
$ Since, n \in \mathbb{N}, n \not\in n$
By contradiction $\{n\}\setminus n \not= \emptyset$
Therefore, $\{n\}\setminus n = \{n\} \implies\{n\}  \subset \emptyset$
$\implies n \in\emptyset \implies false$
By contradiction $ n \bigcup \{n\} \not\subset n$∎
But, is there any set $X: X\in X$ is true. In particular are the natural numbers in the natural numbers given that they're infinite.
 A: 
But, is there any set $X:X\in X$ is true.

No (at least not in normal set theories, others exist). This is important, and essentially the reason why axiomatic set theory exists: allowing things like this makes incredibly weird things happen. This is called well-foundedness (and it's a direct consequence of the Axiom of Regularity in ZF, the most common set theory). 

In particular are the natural numbers in the natural numbers given that they're infinite.

No. If you want to define natural numbers as sets, the usual definition is that $0 = \emptyset$, and $n = \{n - 1\} \cup (n - 1)$. In particular, all such sets are finite, so $\mathbb{N}\not\in\mathbb{N}$. 
A: There is nothing inherently wrong with sets $X$ satisfying $X\in X$. They can be reasoned about just like any other set (which is all we can ask for, I guess).
However, one of the axioms in ZF set theory (which is the most commonly used theory, especially if you count it together with ZFC) is called the axiom of regularity. It implies that there are no such sets.
So there is nothing a priori problematic with such sets. They just never occur in the standard theory.
