$\text{DSPACE}(O(1))=\text{REG}$ Proof?
Suppose you have a Turing machine with an $n$ sized tape with $k$ symbols and $s$ states. This system, as a whole, then has $s\cdot k^n \cdot n$ possible states it can be in ($s$ states, $k^n$ tape contents, $n$ tape head positions).

By definition $k$ and $s$ must be $O(1)$. If $n$ also is however, then the entire expression $s \cdot k^n \cdot n$ is a constant, and we can construct a finite state machine with that many states that has exactly the same behavior as the Turing machine.