What is this variant of set cover problem known as?
Here I show that the problem is NP-complete.

We convert a CNF to an instance of your problem as follows.
Suppose that the variables of the CNF are $n$ $x_i$'s and the clauses are $m$ $C_j$'s, where $n&lt;m$.
Let $U=\cup_i (A_i\cup B_i\cup Z_i)$ where all sets in the union are completely disjoint.
In fact, $A_i=\{a_{i,j}\mid x_i\in C_j\}\cup\{a_{i,0}\}$ and $B_i=\{b_{i,j}\mid x_i\in C_j\}\cup\{b_{i,0}\}$, while $Z_i$ is any set of cardinality $k=2n+1$.
Also denote $Z=\cup_i Z_i$ and fix for every $Z_i$ an increasing family of length $k$ inside it, denoted by $Z_{i,l}$ for $l=1..k$.
For every variable $x_i$, we add $2k$ sets to $\mathcal F$, every set of the form $A_i \cup Z_{i,l}$ and $B_i \cup Z_{i,l}$.
For every clause $C_j$, we add one set to $\mathcal F$, which contains $Z$, and for every $x_i\in C_j$ element $\{a_{i,j}\}$ and for every $\bar x_i\in C_j$ element $\{b_{i,j}\}$.

Suppose that the formula is satisfiable and fix a satisfying assignment.
Then pick the $k$ sets of the form $A_i \cup Z_{i,l}$ or $B_i \cup Z_{i,l}$, depending on whether $x_i$ is true or not.
These are $nk$ incremental sets.
Now add the $m$ sets corresponding to the clauses.
These also keep increasing the size, as the clauses are satisfiable.
Finally, we can even add $k$ more sets (one for each variable) to make the sequence cover $U$.

Now suppose that $n(k+1)+m$ sets are put in an incremental sequence.
Notice that at most $k+1$ sets corresponding to $x_i$ can be selected for each $x_i$.
Thus, if there are no clause sets in the incremental sequence, at most $n(k+1)$ can be selected, which is too few.
Notice that as soon as a clause set is selected, we can pick at most two sets corresponding to each $x_i$, a total of at most $2n$ sets.
Therefore, we have to pick at least $n(k-1)$ variable sets before any clause set is picked.
But as we can pick at most $k+1$ for each $x_i$, this means that for each we have picked at least $1$, as $k=2n+1$.
This determines the "value" of the variable, thus we can pick only "true" clauses.

Update: Changed value of $k$ from $n$ to $2n+1$ as pointed out by Marzio.