I will use binary. I claim to have a bijection $f \colon \mathcal{N} \to \left[ 0, 1 \right)$ where $\mathcal{N}$ is the set of natural numbers $\left\{ 0, 1, 10, 11, 100, \dotsc \right\}$ as follows:
$$ \begin{array}{l|l} x \in \mathcal{N} & f(x) \in \left[ 0, 1 \right)\\\hline 0 & 0.0 \\ 1 & 0.1 \\ 10 & 0.01 \\ 11 & 0.11 \\ 100 & 0.001 \\ 101 & 0.011 \\ 110 & 0.101 \\ 111 & 0.111 \\ 1000 & 0.0001 \\ \hspace{.6em}\vdots & \hspace{.9em}\vdots \end{array} $$
It's basically the shortlex order.
I claim to have everything on this list. Diagonalization? Go ahead! Constructing a number that differs from $f(i + 1)$ in its $i$-th fractional digit yields $0.11111\!\ldots = 0.\overline{1}$ which is $1$, and that should not be on the list anyway!
Why does this not work? I think it has something to do with integers supposedly not having infinite digits, while reals supposedly do.