Another version of the online set cover problem?
A simple adversary argument shows that you must take every set you are offered, unless it's entirely covered by elements already taken. Indeed, otherwise there must be some element $x$ which hasn't appeared before; perhaps it will never appear again, so you must take the set.

Consider now the following instance:
$$
\{x_1\},\{x_2,\},\ldots,\{x_{m-1}\},\{x_1,\ldots,x_m\}.
$$
The algorithm takes all $m$ sets, but the last one would have sufficed. This shows that the approximation ratio can be as bad as $m$, the number of sets.