Problem: Alice and Bob play a game on a $6$ by $6$ grid.
On his turn, a player chooses a rational number not yet appearing in the grid
and writes it in an empty square of the grid.
Alice goes first and then the players alternate.
When all squares have numbers written in them, in each row,
the square with the greatest number in that row is colored black.
Alice wins if he can then draw a line from the top of the grid to the bottom of the grid
that stays in black squares, and Bob wins if he can't.
(If two squares share a vertex,
Alice can draw a line from one to the other that stays in those two squares.)
Find, with proof, a winning strategy for one of the players.

Solution: Bob can win.
Label the first two rows as follows:
\[
  \begin{bmatrix}
    a & b & c & d & e & f \\
    d' & e' & f' & a' & b' & c'
  \end{bmatrix}
\]
These twelve boxes thus come in six \emph{pairs}, $(a, a')$, $(b, b')$ and so on.

\begin{claim*}
  Bob can ensure that the order relation
  of the labels is the same between the two rows,
  meaning that $a < b$ if and only if $a' < b'$, and so on.
\end{claim*}
\begin{proof}
  If Alice plays $q$ in some box in the first two rows, then Bob can plays $q + \eps$
  in the corresponding box in the same pair, for some sufficiently small $\eps$
  (in terms of the existing numbers).

  When Alice writes a number in any other row, Bob writes anywhere in rows $3$ to $6$.
\end{proof}

Under this strategy the black squares in the first two rows will be a pair
and therefore will not touch, so Bob wins.