@hexomino's answer is correct and well-reasoned, as always. Here's another approach, which to me feels much more.. "axe-to-the-head" is what I'd call it in my native language, so I thought it might be interesting enough to warrant posting.
Lower bound: (This is what @trolley813's encrypted comment is saying.)
must be enough, and also required.
white has two unavoidable captures coming up: on white's next move, the
Black can prevent only one of those moves, so
the first objective is completed.
Continuing from there, white will play
next. It threatens black's F-pawn, and also the queen's rook or some other piece in front of it. All those pieces ...
Here's (what I think is) a simpler proof of Albert's lemma than the one in loopy_wall's answer. We'll find either a king-path of 0-squares connecting N and S sides, or a king-path of 1-squares connecting W and E sides. The basic idea is to walk along the boundary between 0-squares and 1-squares until we reach an edge of the board. So here's an example board; ...
Here is a solution with six lines:
It's difficult to tell how I found this, except that I already knew a solution for a 3 by 3 grid with 4 lines, which can be found e.g. here on our sister site Mathematics Stack Exchange. It's also possible that a solution with 5 lines exists.
(By the way, the puzzle is missing the requirement that the lines must be ...
Another proof of Albert's lemma (and one that I believe is much more elegant than the others presented):
I will prove a stronger lemma instead. Namely:
(A proof of the reduced statement follows. This proof is similar to Gareth's answer to the same question, but does not rely on an arbitrary choice of "leftmost".)