Would this be a valid set?
We denote by $fp(\mathbb S)$ the set of all finite subsets of $\mathbb S$, and $ip(\mathbb S)$ the set of all infinite subsets of $\mathbb S$.
$\{x_n \in fp(y) : y \in ip(\mathbb N)\land x_n \notin \{ x_1, x_2, x_3, \dots, x_{n-1} \} \land n \in \mathbb N \} \}$
No, this isn't valid. You reference $x_1,x_2,\ldots$, but haven't defined them previously! Note that set-builder notation does not build a set one step at a time - it tries to build the whole thing at once. You can't use it for a recursive definition. That doesn't mean recursive definitions can't be used - you just can't put them in this format.
Also, in order to introduce things like $n$, you need to say which $n$s you want to consider - "some" $n$? "Every" $n$? "Every third" $n$? You need a quantifier.
You're also being a bit inefficient: saying $y \in \mathbb{I} \wedge \mathbb{I} = ip(\mathbb{N})$ is no different from just saying $y \in ip(\mathbb{N})$.
Ignoring the $x_n \notin \{x_1,x_2,\ldots\}$ part (which doesn't make sense, as I mentioned above), here's what your set is: the set of all finite subsets of infinite subsets of $\mathbb{N}$. This is a weird thing to say, because you could say it much more simply as just "the set of all finite subsets of $\mathbb{N}$", in which case you could have just written $fp(\mathbb{N})$.
To summarize: Your set, as written, is almost valid; but fixing it probably doesn't give the set you wanted to write. If you let me know what set you wanted, I might be able to point you in the right direction.