Understanding Kuratowski finiteness: if $A$ is finite then $A \cup \{x\}$ is finite for all $A$ and $x$ So in this set theory course I am taking, they define finiteness as Kuratowski finiteness.
That is:
Let $X$ be a set. Then $K \subseteq \mathcal{P}(X)$ is Kuratowski inductive if


*

*$\emptyset \in K$

*If $E \in K$ and $x \in X$ then $E \cup \{x\} \in K$ 


We now say that a set $X$ is finite if $X$ belongs to every Kuratowski inductive subsets of $\mathcal{P}(X)$.
So in general I'm really struggling to see what this is mean to mean, or I guess the intuition behind this definition. I'm assuming its something to do with the $K$ sets becoming very large by induction and the finite set needing to eventually belong to it. 
I've also been given a list of (I guess intuitive) properties about finiteness but I don't really know whats going on with the proofs.
For example we try and show that if $A$ is finite then $A \cup \{x\}$ is also finite.
The proof in the notes does this by considering a $K \subset \mathcal{P}(A \cup \{x\})$ and then defines a new set called $K*$ such that $K* = K \cap \mathcal{P}(A)$. They then try and prove that $K*$ is Kuratowski inductive. I'm not really getting that the strategy here is and why this is needed for the proof.
General intuition and motivations behind the definition would also be really appreciated! Thanks
 A: Essentially, it's saying that the collection of Kuratowski finite subsets of $X$ is the smallest set $K$ which contains $\emptyset$ and is closed under the operation of adjoining a single element.  The definition is just forming this collection by forming the intersection of all sets which satisfy these generator and closure conditions.  Then, you define a set $X$ to be (absolutely) Kuratowski finite if $X$ is a Kuratowski finite subset of itself.
For intuition, you would usually think of a Kuratowski finite set as representing something built up from below by starting with $\emptyset$ and adjoining single elements one at a time.  This would end up with a set like $\{ x_1, x_2, \ldots, x_n \}$ for $x_1, \ldots, x_n \in X$ for some $n \in \mathbb{N}$; but the definition as stated avoids needing to assume the existence of a set of natural numbers $\mathbb{N}$.
To prove the statement $S \in K \Rightarrow S \cup \{ x \} \in K$, the outline of the proof would be to show that since $S$ is in every Kuratowski inductive set $K'$, so is $S \cup \{ x \}$, so $S \cup \{ x \}$ is also in the intersection of all Kuratowski inductive sets.
Incidentally, the collection of Kuratowski finite subsets of $X$ also has an induction principle: suppose $\phi$ is a condition on subsets of $X$.  Then if $\phi(\emptyset)$, and whenever $\phi(S)$ and $x \in S$ then $\phi(S \cup \{x\})$, then you can conclude $\phi(S)$ for all Kuratowski finite $S$.  With the definition given, this is just a reformulation of the fact that $\{ S \subseteq X : \phi(S) \}$ is Kuratowski inductive.
So, another step in the proof of your original statement would be: prove that if $S \subseteq X \cap Y$, then $S$ is Kuratowski finite as a subset of $X$ if and only if it is Kuratowski finite as a subset of $Y$.  By symmetry, it's sufficient to prove one direction.  So, let $\phi$ be the condition on subsets of $X$: $S \subseteq Y \Rightarrow S$ is Kuratowski finite as a subset of $Y$.  (Or, if it helps, you might think of $\phi$ as: either $S \not\subseteq Y$ or $S$ is Kuratowski finite as a subset of $Y$.)  It should be easy to check $\phi$ satisfies the hypotheses of the inductive principle, which concludes the proof.
Then, as the final step, if $A$ is (absolutely) Kuratowski finite, then $A$ is a Kuratowski finite subset of itself, so it is also a Kuratowski finite subset of $A \cup \{ x \}$.  Combining this with a previous statement, this implies $A \cup \{ x \}$ is also a Kuratowski finite subset of $A \cup \{ x \}$, so $A \cup \{ x \}$ is absolutely Kuratowski finite.
For a simpler example of the same sort of definition, you might consider this definition of even numbers in $\mathbb{N}$: a subset $S$ of $\mathbb{N}$ will be said to be $(0, \cdot+2)$-closed if $0 \in S$ and whenever $n \in S$, $n+2 \in S$.  Then $n \in \mathbb{N}$ is even if $n$ is in all $(0, \cdot + 2)$-closed subsets of $\mathbb{N}$.  The intuitive point of view from below says that this means an even number can be gotten by starting from 0 and adding 2 a finite number of times.
Or, if you're familiar with some abstract algebra, you can treat this as similar to defining the subgroup of $G$ generated by a subset $S \subseteq G$ as being the intersection of all subgroups $H$ of $G$ such that $S \subseteq H$.  The point of view from below says that an element of the subgroup of $G$ can be built up from $e$ and elements of $S$ by taking products and inverses in a finite number of steps.
