I first had an idea of splitting (0, 1) into parts of ($\frac{1}{2^n}$,$\frac {1}{2^{n+1}}$ and mapping these to (2, 3) ∪ (4, 5) ∪ ... and the remaining {1/2, 1/4, 1/8, ...} to be mapped to (0, 1), but I can't seem to write it as a proof.
Can someone please guide me on how to do so, or how else to approach the problem?
On
You can map $(0,1)\mapsto (0,1]$ like this:
Send $\frac 12\to 1$, now there is a hole in $\frac 12$ so send $\frac 14\to\frac 12$, and so on: $\quad\cdots\to \frac 18\to\frac 14\to\frac 12\to 1$
We can write it $\quad\phi(x):\begin{cases}2x& x=2^{-k}& \forall k\in\mathbb N\\x&\text{otherwise}\end{cases}$
Similarly you can map $(n,n+1)\mapsto(n,n+1]$ and now you can use your initial idea to map the union to $(0,1)$, the stitching issue is now solved.
Use a bijection of $\mathbb{R}$ with $(0,1)$ (like $\mathbb{R}\ni x\mapsto\frac{1}{1+e^{x}}\in (0,1)$) and consider the inclusion of $\bigcup_{n} (n,n+1)$ into $\mathbb{R}$. Using that $(0,1)$ is included into the union, the conclusion follows from Schröder-Bernstein theorem.