Let ${^\Bbb N}\{0,1\}$ be the set of infinite sequences of $0$’s and $1$’s, and let $\Sigma$ be the set of finite sequences of $0$’s and $1$’s. For each $\sigma\in{^\Bbb N}\{0,1\}$ let $$S_\sigma=\{s\in\Sigma:s\text{ is an initial segment of }\sigma\}\;;$$ it’s not hard to show that if $\sigma,\tau\in{^\Bbb N}\{0,1\}$, and $\sigma\ne\tau$, then $S_\sigma\cap S_\tau$ is finite.
Finally, $\Sigma$ is countably infinite, so it admits a bijection with $\Bbb N$. With a bit of work one can produce an explicit bijection $h:\Sigma\to\Bbb N$, and then the collection $\big\{h[S_\sigma]:\sigma\in{^\Bbb N}\{0,1\}\big\}$ has the desired properties.
Added: Here’s a pretty nice bijection $h$. If $s=\langle b_0,\dots,b_{n-1}\rangle\in\Sigma$, let $$h(s)=2^n+\sum_{k=0}^{n-1}b_k2^k\;.$$ As $s$ runs over all sequences in $\Sigma$ of length $n$, $\sum_{k=0}^{n-1}b_k2^k$ runs over all non-negative integers in $\{0,\dots,2^n-1\}$, and $h(s)$ runs over $\{2^n,\dots,2\cdot2^n-1\}=\{2^n,\dots,2^{n+1}-1\}$, so $h$ is clearly a bijection from $\Sigma$ to $\Bbb N$. Indeed, given $n\in\Bbb N$, we can calculate $h^{-1}(n)$ as follows.
First let $k=\lfloor \lg n\rfloor$, where $\lg x$ is the binary log; then $2^k\le n<2^{k+1}$. Let $m=n-2^k$; then $0\le m<2^k$, so $m$ has a unique binary representation $$m=\sum_{i=0}^{k-1}b_i2^i\;,$$ where each $b_i\in\{0,1\}$, and $h^{-1}(n)=\langle b_0,\dots,b_{k-1}\rangle$.
If $\sum a_n$ is a conditionally convergent series of real numbers and $\alpha\le\beta$ then there is a rearrangement so that $\liminf s_n=\alpha$ and $\limsup s_n=\beta$.
The proof is the same as the proof that there is a rearrangement converging to $\alpha$. Start with just enough positive terms to give a partial sum larger than $\beta$, then add just enough negative terms to give a partial sum less than $\alpha$, then add more positive terms til you get a partial sum larger than $\beta$ again, etc. See the proof of the rearrangement theorem that you mention for details.
Best Answer
Here you’ll find a proof that the infinite continued fractions with $0$ integer part are precisely the irrationals in $(0,1)$. The map
$$\left(\Bbb Z^+\right)^{\Bbb Z^+}\to(0,1):a\mapsto[0;a_1,a_2,a_3,\dots]$$
is therefore an injection.