We want to partition $\mathbb{N}$ into $A,B$ such that $A$ has no infinite arithmetic progression and $B$ has no infinite arithmetic progression. Here is my attempt. I am looking for possible verification, or perhaps a quicker solution.
We define our sets recursively. We start with $A_1=\{1\}$ and $B_1=\{2\}$. Now assuming we have $A_n$ and $B_n.$ Let $m=\max B_n$ and then we set $$A_{n+1}=A_n\cup\{m+1,m+2,...,m+n+1\},$$ and $$B_{n+1}=B_n\cup\{(m+n+1)+1,(m+n+1)+2,...,(m+n+1)+n+1\}.$$ Inductively $A_n$ and $B_n$ are disjoint for all $n,$ and $$\{1,...,n\}\subseteq A_n\cup B_n.$$ Now we set $$A=\bigcup_{n\in\mathbb{N}}A_n$$ and $$B=\bigcup_{n\in\mathbb{N}}B_n.$$ Then $A\cup B=\mathbb{N},$ and $A,B$ are disjoint, hence they partition $\mathbb{N}.$ The only thing that remains is to show neither has an infinite arithmetic progression. Actually, by symmetry it suffices to show that $A$ does not.
Suppose that $A$ had an infinite arithemetic progression, say $$\{m+kr:r\in\mathbb{N}\}\subseteq A.$$ It follows that there exist an $N\in\mathbb{N}$ such that $\{n\in A:n\geq N\}$ has no gaps greater than or equal to $k.$ Choose such an $N,$ then choose $n\in\mathbb{N}$ such that $\max{A_n}>N$ and $n>k.$ Then $$\max{A_n}+1,...,\max{A_n}+n+1$$ is a gap in $A$ greater than $k.$ Contradiction, hence $A$ has no infinite arithmetic sequence.
Verification: Yes, you have a valid proof.
A shorter solution:
Let $A$ be the set of all positive integers with an odd number of bits in their binary expansion, and $B$ be the complementary set with an even number of bits. $A=\{1,4,5,6,7,17,\cdots\}$ and $B=\{2,3,8,9,10,11,\cdots\}$. An integer $x$ is in $A$ if $2^{2k}\le x < 2^{2k+1}$ for some $k$, and it's in $B$ if $2^{2k+1}\le x < 2^{2k+2}$.
Given an increasing arithmetic progression $x_n = c + dn$, choose some $k$ such that $c,d < 2^{2k}$. Then, for $n=\left\lceil\frac{2^{2k}-c}{d}\right\rceil$, $2^{2k} \le x_n < 2^{2k}+d < 2^{2k+1}$ and $x_n\in A$. Similarly, for $n=\left\lceil\frac{2^{2k+1}-c}{d}\right\rceil$, $2^{2k+1} \le x_n < 2^{2k+1}+d < 2^{2k+2}$ and $x_n\in B$. Every infinite arithmetic progression contains members of both sets, so none is contained in one of them.
Being explicit with the construction has its advantages.