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?

It is consistent with $ZFC$ that there exists a model of $ZF$ with an infinite Dedekind-finite set. 
Let $X$ be such a set. Let $X_i = \omega$ for all $i \in X$. 
$X$ is infinite Dedekind-finite, it is uncountable. Clearly $\bigcup_{i \in X} X_i = \omega$. Since $X$ is infinite Dedeind-finite, $X$ has no countable subsets, thus for all $J \subseteq X$ such that $|J| = \aleph_0$, $\bigcup_{i \in J} X_i \subsetneq \omega$.

If you weaken $|J| = \aleph_0$ to $|J| \leq \aleph_0$, there still is a sequence (though not the example above). 
Let $X$ be an infinite Dedekind-finite set. As bof mentioned, let $S$ be the set of finite sequences of $X$ with all distinct terms. $S$ is still an infinite Dedekind-finite set. The cardinality function denoted $f$ is a surjection $f : S \rightarrow \omega$. Let $X_s = \{f(s)\}$. This sequence would work. 
