A trivial way to check $C \subset S$ is by searching over $S$ for each element in $C$. First, check if $C$ has at most $n$ elements, which can be done in $O(n)$ (otherwise, the answer is no). Then, for each element in $C$, compare it to all of the elements in $S$ to see if there is a match ($O(n)$ operations per element, and $O(n)$ elements, so this is $O(n^2)$). If there isn't a match for some element, you say no. If all elements have a match, then you say yes. So, this can be done in $O(n^2) + O(n) = O(n^2)$.
You don't need to have the best possible algorithm to show the existence of a P-verifier, just something which you can show that does it in polynomial time (which this does).