I think I spotted a problem with the above proof by Scott. If P=SPACE($n$) implied that there's an algorithm to simulate an n-space Turing machine in (say) $n^c$ time, for some constant $c$, it would also imply that there's an algorithm in $n^c$ time to simulate all polynomial time Turing machines, and that P=TIME($n^c$), which contradicts the Time Hierarchy Theorem. But this problem can be fixed, of course:

Suppose that P=SPACE($n$). For contradiction with the Space Hierarchy Theorem, we'd like to show that every language in SPACE($n^2$) can be reduced in polynomial time to SPACE($n$). We will do this using the padding argument.

Let $L$ ∈ SPACE($n^2$), $M$ a Turing machine with space complexity $n^2$ for the language $L$. Let $L'$ be the “padded” version of $L$ defined as the set of words of the form $x01^{|x|^2}$ for each $x ∈ L$. Now $L'$ is in SPACE($n$) because for an input of size $n$, no more than $\sqrt{n}$ characters are the original input and we can run the simulation of the original machine $M$ using space $n$.

A reduction from $L$ to $L'$ can be easily done in polynomial time, hence P=SPACE($n$)=SPACE($n^2$), and we get the desired contradiction with the Space Hierarchy Theorem.