Writing uncountable set as union of finite sets $I_j$ with $I_j \subseteq I_{j+1}$ Let $I$ be an uncountable set. I would like to prove that it can be written as a union 
$$I=\bigcup_{j\in J} I_j,$$
with $I_j$ finite and $I_j \subseteq I_{j+1}$ for all $j \in J$. $J$ can be any set.
There are probably some axioms and theorems in set theory which help proving this. However, I haven't got much knowledge in this area.
I am grateful for any kind of help.
 A: As nombre comments, your question is unclear. 
Depending what exactly you mean, the result you are trying to prove is false. One natural interpretation of your claim is:

For any uncountable $I$, we can write $I=\bigcup_{j\in\mathbb{N}}I_j$ with $I_j$ finite and $I_j\subseteq I_{j+1}$.

This is clearly false, since any $I$ which can be so written is a countable union of finite sets and hence countable.
One might reasonably try to rephrase this using ordinals:

For any uncountable $I$, we can write $I=\bigcup_{j\in\alpha}I_j$ for some ordinal $\alpha$, with $I_j$ finite and $I_j\subseteq I_{j+1}$.

This is now true; however, note that the picture gets messed up at limit levels. For instance, if $\alpha=\omega+1$, there is no reason that $I_\omega$ has to contain, say $I_5$. This feels a bit weird . . .
. . . and suggests the following formulation:

For any uncountable $I$, we can write $I=\bigcup_{j\in\alpha}I_j$ for some ordinal $\alpha$, with $I_j$ finite and $I_j\subseteq I_{k}$ whenever $j<k<\alpha$.

But this is false again. Indeed, the more general statement

For any uncountable $I$, we can write $I=\bigcup_{j\in J}I_j$ for some linearly ordered set $J$, with $I_j$ finite and $I_j\subseteq I_k$ whenever $j<_J k$

is false. To see this, consider the following preorder $\le_I$ on $I$: $$x\le_Iy\iff \forall j(y\in I_j\implies x\in I_j).$$ That is, $x\le_I y$ if $x$ "enters" $I$ no later than $y$ does. (Note that this of course depends on the decomposition $I=\bigcup I_j$; we're assuming such a decomposition exists, for a contradiction.) Let $\downarrow(x)=\{z: z\le_Ix\}$, let $\equiv_I$ be the equivalence relation gotten from $\le_I$ ($a\equiv_I b\iff a\le_Ib$ and $b\le_I a$), and let $I/\equiv_I$ denote the set of $\le_I$-equivalence classes (note that under $\le_I$, $I/\equiv_I$ is a genuine linear order, whereas $I$ was just a partial order).
Here's the problem: since each $I_j$ is finite, we have $\downarrow(x)$ is finite for every $x\in I$. In particular, this means that the linear order $I/\equiv_I$ is a well-ordering (= no infinite descending chains) in which every element has finitely many predecessors. By a nice exercise about well-orderings, this means there are at most countably many $\equiv_I$-equivalence classes, but since each $\equiv_I$-class is finite, this means $I$ is a countable union of finite sets, so countable.
Finally, there's one more version of your claim to consider: the partially ordered version.

For any uncountable $I$, there is a partial order $P$ such that $I=\bigcup_{p\in P} I_p$ with each $I_p$ finite and $I_p\subseteq I_q$ whenever $p\le_P q$.

This is trivially true: take $P=I$ with the discrete partial order, and $I_p=\{p\}$. So we probably want to assume $P$ is directed: for any $p_0, p_1\in P$, there is a $q\in P$ with $q\ge_P p_0, p_1$. The claim remains true with this demand: take $P$ to be the set of finite subsets of $I$, ordered by inclusion, with $I_p=p$.
