I need to see it visually.
Reduce the problem to a simpler one.
Take a 3x3 board with black and white stones as indicated in the problem statement.
(Think tic-tac-toe board with black in 1 and 9 and white in 3 and 7.)
Bob makes the first move which is to move a black stone next to one of Alice's white stones and reducing Alice's possible moves to 2, 6, and 8 as indicated here.
It doesn't matter which stone moves or to which point, because they are all reflections of each other. So, Alice's first move it always from this position and she has 3 choices: 2, 6, or 8.
If Alice chooses to move 7 to 8 and Bob does not move to 5,
B|2|W 1|2|W 1|2|W 1|2|W 1|x|W
4|5|6 -> B|5|6 -> B|5|6 ?> B|5|B -> B|W|B
W|8|B W|8|B 7|W|B 7|W|9 7|8|9
B|2|W 1|2|W 1|2|W B|2|W B|x|W
4|5|6 -> B|5|6 -> B|5|6 ?> 4|5|B -> 4|W|x
W|8|B W|8|B 7|W|B 7|W|9 7|8|B
Alice will move 8 to 5, providing at least one move that cannot be blocked. So Bob must move to 5.
B|2|W 1|2|W 1|2|W 1|2|W
4|5|6 -> B|5|6 -> B|B|6 -> 4|B|6
W|8|B W|8|B 7|W|B 7|W|B
Alice then moves 3 to 6 reducing the distance between the four stones to the minimum.
At this point Bob must move the black stone at 5 to either 2 or 4 allowing Alice to move either of her stones into the vacated point for the win.
Other possible moves for Alice yield the same results. As long as Alice always moves one of her stones towards the other, no matter how Bob blocks, eventually the stones will converge to minimal distance between the 4 stones and Bob will be forced to move away allowing Alice to move into the vacated point and take the win.
Expand to a 19x19 board and the results will be the same. Go Alice!