Counterexample to greedy solution for set cover problem
Usually the set cover problem is formulated such that the quantity to be minimized is the number of sets one picks rather than the sum of the numbers of elements of the sets picked.
Unfortunately, this means that your example does not prove that the given greedy algorithm fails to find optimal solutions as the cover it produces consists of 3 sets, which is minimal for your instance.
A valid (and perhaps minimal?) counterexample would be $\mathcal U = \{1, 2, 3, 4, 5, 6\}$ with $\mathcal S =\{\{1, 2, 3\}, \{4, 5, 6\}, \{2, 3, 4, 5\}\}$, which works similarly to the one you found.
The minimal cover consists only of the first two sets (these two must be in every cover since there is no other way to account for $1$ and $6$ otherwise) whereas the greedy algorithm will choose the last set first (since it covers 4 rather than 3 elements) and hence picks all sets, leading to a cover of size 3 which is not optimal.