Can a perfect matching always be found by a picking sequence?
Yes, it can. This can be proved by mathematical induction on $n$.

Denote by $U$ the set of people. For a subset of people $P$, denote by $N(P)$ the neighborhood of $P$, that is the set of items that at least one person in $P$ likes. By Hall's marriage theorem, $|P|\le |N(P)|$ for all $P$.

If $|P|&lt; |N(P)|$ for all $P\subsetneq U$, then we can let an arbitrary person pick first. After his picking, $N(P)$ is decreased by at most 1 for all $P$, so $|P|\le |N(P)|$ holds for all $P$ in the remaining graph. Therefore there is still a perfect matching in the remaining graph by Hall's marriage theorem, and there is a valid picking sequence by inductive assumption.

If there exists $P\subsetneq U$ such that $|P|=|N(P)|$, then there is a perfect matching in the subgraph induced by $P\cup N(P)$, and by inductive assumption there is a valid picking sequence for people in $P$ in this subgraph. We can apply this picking sequence in our orignial graph first, and the result is the same, that is, exactly all items in $N(P)$ are picked. Note there is still a perfect matching in the remaining graph, so we can again use the inductive assumption to complete the picking sequence.