A basic set theory question on the decompositions of the set of all irrational numbers $\mathbb I$ I am thinking about:

How to construct (or just prove that at least one exists, if it exists at all?) some countably infinite family of sets $I_i$ such that $\bigcup_{i=1}^{+ \infty}I_i=\mathbb I$ and, every $I_i$ is uncountable, and $I_i \cap I_j=J_{ij}$ is uncountable, and if $i \neq j$ then both $I_i \setminus I_j$ and $I_j \setminus I_i$ are uncountable.

I feel like something is unsatisfiable here, meaning that there are no such families, and that there also should be a very simple proof of that, which I miss.
 A: Since $\bigcup \{J_{ij}:(i,j) \in \mathbb N \times \mathbb N \}=\mathbb I$ and 
$\bigcup_{i=1}^{+ \infty}I_i=\mathbb I$ then $((\bigcup_{i=1}^{+ \infty}I_i) \cup (\bigcup_{i=1}^{+ \infty}I_i)) \setminus \bigcup \{J_{ij}:(i,j) \in \mathbb N \times \mathbb N \}=\mathbb I \setminus \mathbb I= \emptyset$ but $((\bigcup_{i=1}^{+ \infty}I_i) \cup (\bigcup_{i=1}^{+ \infty}I_i)) \setminus \bigcup \{J_{ij}:(i,j) \in \mathbb N \times \mathbb N \}= \bigcup (I_j \setminus I_i)\cup(I_i\setminus I_j))=W$ so $W= \emptyset$ but $\emptyset$ is not uncountable so such family does not exist!
Edit: As mentioned in the comments, this approach does not solve this problem if done like this, so this is not a solution.
A: Take, for each $n\in\Bbb{N}$, the uncountable set $I_n = \Bbb{I} \setminus [n, n+1[$. Their union is $\Bbb{I}$. Furthermore, if $i \not= j$, then the sets $I_i \cap I_j = \Bbb{I} \cap ([i, i + 1[ \cup [j,j+1[)$, $I_i \setminus I_j = \Bbb{I} \cap [j, j+1[$ and $I_j \setminus I_i = \Bbb{I} \cap [i, i+1[$ are uncountable.
