Condidering the proof, NLOGSPACE$\subset$PSPACE
I wrote following proof:
NLOGSPACE = NSPACE$(\log n)$ $\hspace{15pt} \because$ Definition of NLOGSPACE
NSPACE$(\log n)$ $\subseteq$ DSPACE$(\log^2 n)$ $\hspace{15pt} \because$ Theorem(1)
We let $m = \log n$, then,
DSPACE$(m^2)$ $\subset$PSPACE $\hspace{15pt} \because$ Definition(2)
Therefore, NLOGSPACE$\subset$PSPACE.
Theorem(1): If $S$ is a space constructible function and $S(n) \ge \log n$, then NSPACE$(S(n)) \subseteq$ DSPACE$(S^2(n))$.
definition(2): PSPACE = DSPACE$(n)$$\cup$DSPACE$(n^2)$$\cup$DSPACE$(n^3)$$\cup$...$\cup$DSPACE$(n^k)$$\cup$...
But I am still not that sure if that "We let $m = \log n$" is ok, is this proving NLOGSPACE$\subset$PSPACE?