NMO Selection Tests for BMO and IMO

The answer is in the affirmative. It is sufficient to prove that there exists a positive number $d$ such that, by Ann providing Bob suitable numbers, she eventually forces the $x$-coordinate of the point increase by at least $d$. Then, using $d$ over and over again, she successively increases the $x$-coordinate of the point by at least $2d, 3d, \dots$, thus making it as far away from $A$ as needed.

We now prove that $d = 1/2^{99}$ fits the bill. Ann first provides $1/2^{99}$, then $1/2^{98}$, $1/2^{97}$ and so on and so forth until Bob is forced to make his first move rightward. This latter occurs at step 100 at latest, by the restriction condition on Bob's choices. Let Bob's first move rightward occur at step $k \le 100$. Clearly, if $k = 1$, the case is settled, so let $k \ge 2$.

The first $k-1$ moves are leftward, upward or downward. Any leftward move decreases the $x$-coordinate of the point by the corresponding number Ann provided; upward and downward moves do not change it.

During the first $k$ steps, Ann successively provides the numbers
$$\frac{1}{2^{99}}, \frac{1}{2^{98}}, \dots, \frac{1}{2^{100-(k-1)}} \text{ and } \frac{1}{2^{100-k}}$$
The first $k-1$ moves decrease the $x$-coordinate by at most
$$\frac{1}{2^{99}} + \frac{1}{2^{98}} + \dots + \frac{1}{2^{100-(k-1)}} = \frac{1}{2^{100-k}} - \frac{1}{2^{99}},$$
and the $k$-th increases it by $1/2^{100-k}$. Consequently, the overall variation of the $x$-coordinate is at least
$$- \left( \frac{1}{2^{100-k}} - \frac{1}{2^{99}} \right) + \frac{1}{2^{100-k}} = \frac{1}{2^{99}},$$
as needed. This ends the proof.