SOLVED
Let $A=\{a_1,a_2,\ldots\}\subset \mathbb{R}^d$ be an infinite sequence such that $a_{i+1}-a_i$ is a positive unit vector (i.e. is of the form $(0,0,\ldots,1,0,\ldots,0)$). For which $d$ must $A$ contain a three-term arithmetic progression?
This is true for $d\leq 3$ and false for $d\geq 4$.

This problem is equivalent to one on 'abelian squares' (see [231]). In particular $A$ can be interpreted as an infinite string over an alphabet with $d$ letters (each letter describining which of the $d$ possible steps is taken at each point). An abelian square in a string $s$ is a pair of consecutive blocks $x$ and $y$ appearing in $s$ such that $y$ is a permutation of $x$. The connection comes from the observation that $p,q,r\in A\subset \mathbb{R}^d$ form a three-term arithmetic progression if and only if the string corresponding to the steps from $p$ to $q$ is a permutation of the string corresponding to the steps from $q$ to $r$.

This problem is therefore equivalent to asking for which $d$ there exists an infinite string over $\{1,\ldots,d\}$ with no abelian squares. It is easy to check that in fact any finite string of length $7$ over $\{1,2,3\}$ contains an abelian square.

An infinite string without abelian squares was constructed when $d=4$ by Keränen [Ke92]. We refer to a recent survey by Fici and Puzynina [FiPu23] for more background and related results, and a blog post by Renan for an entertaining and educational discussion.

Additional thanks to: Boris Alexeev and Dustin Mixon