Can the natural number have an uncountable set of subsets? Let $\mathbb{N}$ be the set of natural numbers. Let $X_{i},i\in I$ be an uncountable sequence of subsets such that
$$
\bigcup_{i\in I}X_{i}=\mathbb{N}
$$ 
and
$$
\bigcup_{i\in J}X_{i}\subsetneq \mathbb{N}, \forall J\subset I, |J|=\aleph_{0}
$$
where the inclusion is strict. 
My questions are:
1) Is this possible? (in the light of the answer, I want it in absence of AC)
2) How "bad" is this if it is possible? Can we have even worse examples of this sort?
3) For generalization, if $|X|$ has certain cardinality $\gamma$. Is it possible for us to construct a sequence of subsets with cardinality $\alpha>\gamma$ and satisfies the above two similar conditions?

Using Choice freely, for every natural number $n$ we pick an $i_n$ such that $n\in X_{i_n}$. Then the set of indices picked has cardinality $\le \aleph_0$, and the union of the $X_{i_n}$ is $\mathbb{N}$. Thus, at least in the presence of Choice, the answer to "is this possible" is no. 
