Proving that $\mathrm{SPACE}(o(\log\log n)) = \mathrm{SPACE}(O(1))$?
You're basically there. If the machine uses at most $s(n)$&nbsp;tape cells for inputs of length&nbsp;$n$, then it can't visit more than $C(n)=|Q||\Sigma|^{s(n)}s(n)$ different configurations (possible states times possible tape contents times possible head positions).  Since the machine decides a language, it must halt for all inputs, which means that it can't repeat a configuration, since a deterministic machine that repeats a configuration must loop forever. Therefore, by the pigeonhole principle, it can't run for more than $C(n)$ steps on any input of length&nbsp;$n$.

If $s\in o(\log\log n)$ then $C(n) = o(n)$ so the machine terminates after fewer than $n$&nbsp;steps for all sufficiently large&nbsp;$n$. Intuitively, the machine can't stop in, say, $\sqrt{n}$ steps because, in that many steps, it hasn't seen the whole input, so it doesn't even know what $n$&nbsp;is.

More formally, let $t(n)$ be the greatest number of steps taken by the machine on inputs of length&nbsp;$n$. Let $w$&nbsp;be an input long enough that $t(n)&lt;n$ for all $n\geq |w|$, and choose&nbsp;$w$ so that the machine runs for $t(|w|)$ steps on this input.  We know that $t(|w|)\leq C(|w|)&lt;|w|$, so the machine must terminate before it has time to read the whole input. Because it doesn't read the whole input, its execution on input&nbsp;$wz$ must be the same as its execution on&nbsp;$w$, for any string&nbsp;$z$. Therefore $t(|wz|)=t(|w|)$ for all strings&nbsp;$z$, so the function&nbsp;$t$ is constant for all inputs greater than length&nbsp;$|w|$.

Because it halts in a constant number of steps, it only uses a constant number of tape cells, so the language it decides is in $\mathrm{SPACE}(O(1))$.