Consider the function $d$ mapping the Intervals $I = \{[a,b) \mid a<b \}$ to its power set defined by $$d([a,b)) := \{ [a+(b-a)2^{-(k+1)},a+(b-a) 2^-k \mid k=0,1,2,3,\ldots \}$$
So for example $d([0,1)) = \{ [1/2,1), [1/4,1/2), [1/8,1/4), [1/16,1/8), \ldots\}$
Consider the $k$-th iteration
$$d^k(I) = \bigcup_{J \in d(I)} d^{k-1}(J)$$
(where $d^0 = id$)
So $d^k([a,b))$ is again a set of intervals of the form $[u,v)$
Now we decompose the set $[0,\infty) = \bigcup_{I \in Z}I$ as follows:
$$Z = \bigcup_{k \in \mathbb N} d^k([k-1,k))$$
Now the question is, is $Z$ countably infinite, or is it larger?
EDIT: Obvisouly $d(I)$ is countable. But $d^2(I)$ basically contains countably infinite times a countably infinity, so we could say $|d(I)| "=" |\mathbb N|$, and $|d^2(I)| "=" |\mathbb N|\cdot |\mathbb N|$ and so $|d^k(I)| "=" |\mathbb N|^k$. (I do not want to write equality signs as I don't know whether this notation actually is meaningful.) So you could say that $|Z| "=" \sum_{k=1}^\infty |\mathbb N|^k$.
Hint: prove by induction on $k$ that for each interval $I$ the set $d^k(I)$ is countable. Then use the fact that a countable union of countable sets is countable.
E.g. $\displaystyle d^2(I) = \bigcup_{J \in d(I)} d(J)$. Each $d(J)$ is countable and $d(I)$ is countable, so this is a union of countably many countable sets, so it's countable.