Arithmetic Progressions in slowly oscillating sequences An infinite sequence ($a_0$, $a_1$, ...) is such that the absolute value of the difference between any 2 consecutive terms is equal to $1$. Is there a length-8 subsequence such that the terms are equally spaced on the original sequence and the terms form an arithmetic sequence from left to right?
Clarifications:
1. The Common difference can be negative or 0
Example:
the sequence 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 9, 8, 9, 10
works because the 3rd term is 3, 6th term is 4, 9th term is 5, ..., 24th term is 10. 
and 3rd, 6th, ..., 24th terms are equally spaced. They also form an arithmetic sequence.
I am thinking about Szekeres theorem but idk if that would work
EDIT: I was able to show it for $n=$. I am actually more interested in indefinitely long ones. But hey, it could be that one can construct something that an indefinitely long one will never happen.

Far from a full answer, but I have some (hopefully) new information. Length $4$ equally spaced, AP subsequences can be found from all finite $(a_n)_{n=1}^N$ with $N>10$ and $\forall n(|a_{n+1}-a_n|=1)$. This can very easily be brute forced, as there exist only $2^N$ distinct length $N$ sequences which obey the absolute value condition. That comes out to only $2048$ distinct sequences which require checking.
Here is an example of a length $10$ sequence which does not contain any length $4$, equally spaced, AP subsequences. However, appending either $a_{10}+1$ or $a_{10}-1$ to it will negate this.
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So I thought that there is probably some finite length $N$ after which every such sequence will contain length $8$, equally spaced, AP subsequences. Turns out that even for length $5$, $N$ would have to be greater than $32$. There are $2^{32}$ distinct sequences, and filtering out those sequences where length $5$ APs have already been found, over $3$ million sequences are left. This was when I got a memory error.
Perhaps some of you out there with better hardware and/or programming prowess (or just more time) could brute force the solution, if there is indeed such a finite $N$. Of course, a positive answer for $k$ will beg the same question for $k+1$ and eventually you will run out of processing power or memory, which is why this is a rather inelegant method of doing it.
