Another beautiful example of a sequence with no infinite progressions is any Sturmian word, which can be understood as the sequence generated by taking some irrational number $\alpha$ and using the set $\Bbb N\cup\alpha\Bbb N$, with the elements of $\Bbb N$ colored red and the elements of $\alpha\Bbb N$ blue, and then putting them in increasing order to induce a coloring of $\Bbb N$. For example, with $\alpha=\varphi\approx 1.618$ we get $$\color{red}{1},\color{blue}{\varphi},\color{red}{2},\color{red}{3},\color{blue}{2\varphi},\color{red}{4},\color{blue}{3\varphi},\color{red}{5},\color{red}{6},\dots$$

which translates to the coloring of $\Bbb N$:

$$\color{red}{1},\color{blue}{2},\color{red}{3},\color{red}{4},\color{blue}{5},\color{red}{6},\color{blue}{7},\color{red}{8},\color{red}{9},\dots$$

Using the characterization of Sturmian words in terms of Beatty sequences, we can prove that it has no infinite APs. Suppose $w_n$ is a Sturmian word, treated as a sequence on $0,1$ representing red and blue. Then there exists an irrational $\alpha>0$ and $x\in\Bbb R$ such that $w_n=\lfloor \alpha n+x\rfloor-\lfloor \alpha(n-1)+x\rfloor$. Since the complement of a Sturmian word is also a Sturmian word, we can assume WLOG that the infinite AP is red, so suppose that $w_{dn+a}=0$ for $n\in\Bbb N$, meaning that $\lfloor \alpha (dn+a)+x\rfloor=\lfloor \alpha (dn+a-1)+x\rfloor$, or by setting $\beta=\alpha d$ and $x'=x+\alpha a$, $$\lfloor \beta n+x'\rfloor=\lfloor \beta n-\alpha+x'\rfloor,$$

which in particular means that there is no integer $m$ in $(\beta n-\alpha+x',\beta n+x')$. This in turn implies that the sequence $\beta n\pmod 1$ (the fractional part of $\beta n$) always misses the nonempty open interval $(-x',\alpha-x')$ (also $\bmod 1$), which contradicts the fact that $\beta n\pmod 1$ is dense (which is itself a good exercise, provable by the pigeonhole principle).