Players $A,B,C,D$ stands in a line.
Players $A, D$ do not move.
round $1:$ player $B$ moves one distance closer to the midpoint of $A$ and $C$
round $2:$ player $C$ moves one distance closer to the midpoint of $B$ and $D$
How should I formulate the problem so to prove that $A,B,C,D$ will be at equidistant at the end of this game?
The question might need some refinement to avoid oscillations around the convergence of $B$ to $\frac{1}{3}$ of $AD$ and $C$ to $\frac{2}{3}$ of $AD$. So it is clear that you don't want to have movements on grid and you may avoid the term "one distance". Why not just stating that $B$ moves to the midpoint of $AC$ and $C$ moves to the midpoint of $BD$? Assume $A$ at $0$ and $D$ at $1$.
By observing that a move of $B$ is limited into the range of $0-0.5$ while $C$ may be at any point on $AD$, and $C$ moves to the segment $0.5-1$ when $B$ is anywhere on $AD$ you may realize that the dependency between $B$ and $C$ moves could lead to convergence. When $B$ and $C$ are limited to the $0-0.5$ and $0.5-1$ ranges you could show that this derives that actually they will be limited to $0.25-0.5$ for $B$ and $0.5-0.75$ for $C$.
Since you are not asking for a solution I will leave it here and hope to be granted the solution:)
The following table demonstrate the ranges convergence for the $B$ and $C$ values:
$$\begin{array}{| 1 | 1| 1| 1| 1| 1| 1| 1| 1| 1| 1|} \hline B_{min} & 0 & 0.00 & 0.25 & 0.25 & 0.31 & 0.31 & 0.33 & 0.33 & 0.33 & 0.33 \\ \hline B_{max} & 1 & 0.50 & 0.50 & 0.38 & 0.38 & 0.34 & 0.34 & 0.34 & 0.34 & 0.33 \\ \hline C_{min} & 0 & 0.50 & 0.50 & 0.63 & 0.63 & 0.66 & 0.66 & 0.66 & 0.66 & 0.67 \\ \hline C_{max} & 1 & 1.00 & 0.75 & 0.75 & 0.69 & 0.69 & 0.67 & 0.67 & 0.67 & 0.67 \\ \hline \end{array}$$