FALSE: every two-colouring of $\mathbb{N}$ contains an infinite monochromatic arithmetic progression

Colour $0$ red. For $n\ge 1$, colour $n$ red when the unique $m$ with $2^m\le n<2^{m+1}$ is even, and blue when $m$ is odd. Thus consecutive dyadic blocks alternate colours, with every power of two assigned to the block beginning there.

Fix $a\in \mathbb{N}$ and $d>0$. For every sufficiently large $m$, let $q_m$ be the least $q$ with $a+qd\ge 2^m$. Minimality gives $a+q_{m}d<2^m+d<2^{m+1}$, so the progression meets the $m$th dyadic block. It therefore meets infinitely many blocks of each parity and contains both colours.

No infinite arithmetic progression is monochromatic in this colouring. This does not contradict [L1], which guarantees arbitrarily long finite progressions only. The displayed universal statement is false.