 Solution

This is an infinite binary expansion with digits in $\{0,1\}$. If the usual expansion of $\alpha$ has infinitely many $1$ digits, list their positions as $b_1<b_2<\cdots$. If it terminates with a final term $2^{-k}$, replace that term bythis also covers every dyadic rational. For $\alpha =1$, use $b_n=n$. Thus every $\alpha \in (0,1]$ has the required form.

The representation by an infinite strictly increasing sequence is unique. If two sequences first differ at exponent $k$, one sum contains $2^{-k}$ and the other does not. The latter's entire possible tail is at most ${\sum }_{j>k}{2}^{-j}=2^{-k}$, while the former has $2^{-k}$ plus its own nonempty tail, so the sums cannot agree.