Show that NP is not equal to SPACE(n)
In order to prove that $\mathsf{SPACE}(n) \not\subseteq \mathsf{NP}$, you need to identify a language in $\mathsf{SPACE}(n)$ which is not in $\mathsf{NP}$. Not every language in $\mathsf{SPACE}(n)$ would work &mdash; for example, every regular language is in $\mathsf{SPACE}(n)$, and all of these are also in $\mathsf{NP}$.

Your argument, as I understand it, is roughly as follows:


Every machine in $\mathsf{SPACE}(n)$ runs in at most exponential time.
Hence every such machine is not in $\mathsf{NP}$.


The second conclusion is not correct, since exponential time is only an upper bound. For example, a DFA uses constant space and does run in polynomial time.

Worse, even if some machine in $\mathsf{SPACE}(n)$ does run in exponential time, it could be that there exists another machine accepting the same language that runs in polynomial time &mdash; especially if we are allowed to use nondeterminism. As an example, it is suspected that $\mathsf{SAT}$ (which is in $\mathsf{SPACE}(n)$) requires exponential time, but it does belong to $\mathsf{NP}$.

The proof outlined in your link is not too hard, given the space hierarchy theorem. This theorem gives you a language $L$ which is in $\mathsf{SPACE}(n^2)$ but not in $\mathsf{SPACE}(n)$. The padded language $L' = \{ 1^{|x|^2}0x : x \in L \}$ does lie in $\mathsf{SPACE}(n)$ (why?), and so if the latter is equal to $\mathsf{NP}$, it follows that $L' \in \mathsf{NP}$. However, $L$ polytime reduces to $L'$, and so it would follow that $L \in \mathsf{NP}$ as well, contradicting $L \notin \mathsf{SPACE}(n)$.