Let $A $ be an arbitrary decision problem in $NP$. I want to determine whether or not any subset of $A$ is also in $NP$. The following simple line of reasoning shows that this is in fact not the case, although I am suspicious of the reasoning.
Let $A = \{0,1\}^*$ be a pathological example of an $NP$-set. Membership of $x$ may be determined trivially of course. Now take a problem $E \notin NP$, then, assuming a common alphabet, $E \subset A$, but $E \notin NP$. Therefore the subsets of $NP$ problems are not necessarily in $NP$, so QED?
I guess this is enough but the result is not really satisfying because we take a set $A$ that technically is in $NP$ but really is in $TIME(1)$. Can anyone give a more general reasoning why we might find sets in $NP$ that have subsets not in $NP$?
Intuitively I can imagine the existence of some $NP$ problem $A$ that has an infinite intersection with some more complicated problem $B$, so that $A \cap B \subset A$ is not in $NP$, but I cannot find a solid argument for this other than my trivial example.
Your proof is fine, and it's the proof that I'd give. But if, for some reason, you prefer to use a complicated NP set $A$, then you could start with any infinite NP set $A$ and observe that (1) it has uncountably many subsets, whereas (2) only countably many sets are in NP. So $A$ must have subsets (in fact lots of them) that are not in NP.) The same argument shows that $A$ has subsets that are not even computable, not even arithmetically definable, not even hyperarithmetical.)