Are there any non-discrete subsets of $\mathbb{R}$ for which this "binary search game" would terminate? Let $X \subseteq \mathbb{R}$ be some subset of the real numbers. We start with two players, Player 1 and Player 2, and each is given a point in $X$, Player $x_1$ and Player 2 $x_2$. The players take turns sequentially. 
On Player 1's turn, Player 1 chooses a point $y \in X$ (the condition that $y \in X$ is crucial) between $x_1^{(n)}$ and $x_2^{(n)}$, i.e. $x_1^{(n)} <y < x_2^{(n)}$, and then forces Player 2 to mover there, i.e. $x_2^{(n+1)} \leftarrow y$. Then on Player 2's turn, Player 2 chooses another point $z \in X$ such that $x_1^{(n)} < z < x_2^{(n+1)}$ and forces Player 1 to move there, i.e. $x_1^{(n+1)} \leftarrow z$. The game ends on Player 1's turn when there does not exist any $x \in X$ such that $x_1^{(n)} < x < x_2^{(n)}$, and Player 1 wins because Player 2 can't force them to move anywhere. Similarly, the game ends on Player 2' turn when there does not exist any $x \in X$ such that $x_1^{(n)} < x < x_2^{(n+1)}$, and Player 2 wins because Player 1 can't force them to move anywhere.

Question: Are there any non-discrete subsets of $\mathbb{R}$ for which this "binary search game" would terminate for every player strategy used by Player 1 and/or Player 2?

In particular, I claim that this would always terminate for $X$ discrete. 
Given $X$ discrete, the game starts by restricting to the space $[x_1, x_2] \cap X$, which would then be compact and discrete, and thus finite. Then termination would follow from a modification of the proof that regular binary search terminates.
Also, it seems like in order for this to terminate, we need that $[x_1, x_2] \cap X$ is finite (please correct me if I am wrong, I have yet to attempt a rigorous proof). Thus if such a non-discrete $X$ exists, it would have to be such that $[x_1, x_2] \cap X$ is finite for every pair $(x_1, x_2) \in \mathbb{R}$.
Note: When I say discrete, I mean in the topological sense that all points are open sets (the discrete topology), not that the set is countable (e.g. the rational numbers are a countable subset of the reals but not a discrete subset).
 A: Let $X=\{m+\frac 1{n}\mid m\in\Bbb Z, n\in\Bbb N\,\}$. This set is not discrete. However, no matter where the players start, if $x_1$ is in some interval $[m,m+1)$, it will tke only finitely many rounds until it is in $[m+1,\infty)$. Hence after still finitely many steps, $x_1$ and $x_2$ are between the same consecutive integers and from then on, it will only take finitely many rounds until the game terminates.

Edit: To incorporate what Noah Schweber wrote in a comment:
Let $X\subseteq \Bbb R$ and assume that for some $\alpha\in\Bbb R$, the sets $X\cap (\alpha,\infty)$ and $-X\cap (-\alpha,\infty)$ are both not well-ordered (under the standard order of the reals). Then there exist non-empty sets $X_1\subseteq X\cap(-\infty,-\alpha)$ without maximum and $X_2\subseteq X\cap(\alpha,\infty)$ without minimum. Thus with respective starting points in these sets, both players can make moves endlessly without ever leaving $X_i$ and hence without ever crossing $\alpha$; the game would not terminate.
We concluder that for any set $X$ with the desired property, for each $\alpha\in\Bbb R$, at least one of $X\cap(\alpha,\infty)$, $-X\cap(-\alpha,\infty)$ must be well-ordered. Now if $X$ is not discrete, it has at least one accumulation point $\xi$. If $\alpha<\xi$ and $X\cap (\alpha,\infty)$ is well-ordered, then $X\cap(\xi,\infty)$ has a minimal element or is empty; at any rate $X\cap(\xi,\xi+\delta)=\emptyset$ for suitable $\delta>0$. Similarly, if $\alpha>\xi$ and $-X\cap(-\alpha,\infty)$ is well-ordered, we conclude that $X\cap(\xi-\delta,\xi)=\emptyset$ for suitable $\delta>0$.  As $\xi$ is an accumulation point, we cannot have both of these properties. Hence we have the following possibilities:


*

*For all $\alpha\in\Bbb R$, $X\cap(\alpha,\infty)$  is well-ordered

*For all $\alpha\in\Bbb R$, $-X\cap(-\alpha,\infty)$  is well-ordered

*For all $\alpha>\xi$, $X\cap(\alpha,\infty)$  is well-ordered; and for all $\alpha<\xi$,  $-X\cap(-\alpha,\infty)$  is well-ordered


The second variant is only a flipped version of the first, and the third variant cannot have a second accumulation point, so boils down to an order type of at mosz $2\omega$.
