Ordered set cover problem: is it NP-hard?
The decision version of your problem is NP-hard (and so, NP-complete), by reduction from SAT.

Let $\varphi$ be a SAT formula on variables $x_1,\ldots,x_n$ and clauses $C_1,\ldots,C_m$. The universe will be $\{x_1,\ldots,x_n,C_1,\ldots,C_m\}$. For every literal $x_i$ or $\bar{x}_i$, there will be a set $S_i$ or $\bar{S}_i$ consisting of $x_i$ together with all clauses satisfied by the literal, in which all elements have unit weight. This instance has a solution of weight at most $n$ if and only if $\varphi$ is satisfiable.

Indeed, if $\varphi$ is satisfiable, then the sets corresponding to a satisfying assignment are a solution of weight $n$, in any order.

Going in the other direction, Suppose that $T_1,\ldots,T_r$ is a solution of weight at most $n$. Any solution must cover $x_1,\ldots,x_n$, so it must contain either $S_i$ or $\bar{S}_i$. If a solution contains both $S_i$ and $\bar{S}_i$, then the one which appears later must have zero cost, that is, its elements must all be covered by preceding sets; so the solution remains a solution even if we remove it. Therefore we can assume that exactly one of $S_i,\bar{S}_i$ appears. The corresponding truth assignment is easily seen to be a satisfying assignment of $\varphi$.