More details on a language decided in $\Theta(\log \log n)$ space
The trick here is to forget about the details as stated in the answer, and think about what would make the claim correct.

If $k$ were known, you could verify that the input is indeed $b(0)\#\cdots\#b(2^k-1)$ as follows: (below, a word is stuff between adjacent sharp signs)


The first word is $0^k$.
For any two contiguous words $x,y$, it holds that $y=x+1$.
The last word is $1^k$.


This can be accomplished in space $O(k)$.

Unfortunately, we don't know $k$ in advance, and even if we did, it's not necessarily the case that $k = O(\log \log n)$ – there could be much less than $2^k-1$ words. Therefore we use a slightly different approach. Let $K$ be the length of the first word. For each $t \leq K$, we verify the following:


The first word ends with $0^t$.
For any two contiguous words $x,y$, the $t$-bit suffixes $x',y'$ satisfy $y'=x'+1$ modulo $2^t$.
The last word ends with $1^t$.


If these tests pass for all $t \leq K$, then we accept the word. The $t$th test can only pass if the input is at least $(t+1)2^t-1$ symbols long and uses only $O(\log t)$ space, and this implies that the algorithm uses $O(\log\log n)$ space.