Your question is not very different from set cover (it would be exactly set cover if you stopped as soon as you found a set containing $x$ rather than keeping going until you have determined $x$) and it's easy to adapt bad instances to set cover to show that the greedy algorithm can ask more questions than optimal by a logarithmic factor.
To see this, suppose that you have $n$ points in the plane, placed at the intersections of $k$ horizontal lines and a larger collection of vertical segments: $\lfloor k/3 \rfloor$ segments through triples of points on disjoint horizontal lines, $\lfloor k/4\rfloor$ segments through quadruples of points on, $\lfloor k/5\rfloor$ segments through five points, etc. Let $Q$ be the sets of points on each of these segments.

The greedy algorithm will begin by querying the vertical segments, in order from longest to shortest, because at each step this choice eliminates more points than taking any horizontal segment. Therefore, it uses a total of $(1-o(1))k\ln k$ tests in the worst case to find $x$. The optimal algorithm will take the horizontal segments first, taking $k$ steps to find a segment containing $x$ and at most $2k$ total to identify $x$. Since $n\approx k^2$, the factor $\tfrac{1}{2}\ln k$ by which greedy is bigger than optimal is $\Theta(\log n).$
(This is a standard counterexample for set cover. Its geometric realization, and the figure, are taken from my manuscript "Forbidden Configurations in Discrete Geometry". Probably you could get the inapproximability ratio up to $\ln n$ by using hard instances for set cover but I think the geometry of this example makes it easier to understand.)
On the other hand, suppose that the optimal strategy for your problem takes $k$ questions in the worst case, and consider the specific sequence of at most $k$ questions that this strategy would follow against a greedy adversary (one that always chooses the answer that leaves the most remaining elements). One of these questions must eliminate at least $n/k$ elements at the time it is asked by the optimal strategy, and would eliminate at least as many elements if it is asked immediately, as the first question. So, at each step, the greedy algorithm also eliminates at least $n'/k$ elements, where $n'$ is the number of remaining elements. Therefore, its total number of steps is $\bigl(1+o(1)\bigr)k\ln k$. That is, the $\ln k$ approximation ratio that one could presumably prove by using worst-case set cover instances in place of the geometric example above is tight.