To prove that a problem $P$ is PSPACE-complete, you need to do two things:
- Show that $P$ can be computed in PSPACE.
- Show that every problem $Q \in $PSPACE reduces to $P$, in the sense that there exists a polytime function $f$ such that $f(x) \in P$ iff $x \in Q$.
As mentioned by dkuper, you forgot about the first part. Your argument for the second part (PSPACE-hardness) unfortunately does not constitute a proof. You need to follow the definition: given an arbitrary $Q \in $PSPACE, produce a polytime function $f$ such that $f(x) \in P$ iff $x \in Q$.
Apart from not being a proof, your argument is also not convincing: there could be some trick that avoids simulating $M$. Such tricks are behind results such as L=SL or NL=coNL in complexity theory. For example, the second result states that given a non-deterministic Turing machine $M$ that runs in logarithmic space, there is a ways to decide non-deterministically in logarithmic space whether a given input is not accepted by $M$. The obvious way to do this is to run all possible execution paths of $M$, but since $M$ could make polynomially many non-deterministic decisions, that would require polynomial space (we need to keep track of all these decisions so that we can run through all of them). Yet there is a trick that enables us to do it non-deterministically using only logarithmic space.
Hint for the second part: Given $Q$, consider a Turing machine $M$ that computes $Q$ in space $O(n^k)$ for some $k$.