Is there an infinite set of finite strings such that no element is a subsequence of another?
Your intuition is correct. Quoting Wikipedia on Higman's lemma:
Higman's lemma states that the set of finite sequences over a finite alphabet, as partially ordered by the subsequence relation, is well-quasi-ordered. That is, if $w_1,w_2,\dots$ is an infinite sequence of words over some fixed finite alphabet, then there exist indices $i\lt j$ such that $w_i$ can be obtained from $w_j$ by deleting some (possibly none) symbols. More generally this remains true when the alphabet is not necessarily finite, but is itself well-quasi-ordered, and the subsequence relation allows the replacement of symbols by earlier symbols in the well-quasi-ordering of labels. This is a special case of the later Kruskal's tree theorem. It is named after Graham Higman, who published it in 1952.
This answer is translated (with small modifications) from here.
$\Sigma$ is a finite alphabet.
$\Sigma^\ast$ is the set of finite strings over $\Sigma$ (Kleene star).
$x\preceq y$ means that $x$ is a subsequence of $y$.
We'll prove that there is no infinite set $S \subseteq \Sigma^\ast$ such that no element of it is a subsequence of another (Higman's lemma).
Assume the thesis is false. Then there is an infinite sequence $x_1, x_2,\ldots$ such that
- $x_i\in\Sigma^\ast$
- $i<j \implies \textit{not} (x_i \preceq x_j) $ (notice that $x_i \succ x_j$ is possible)
From infinite sequences meeting the criteria 1-2 take one that's minimal in the sense that $|x_1|$ is minimal and with $|x_1|$ fixed $|x_2|$ is minimal, etc.
Take an infinite subsequence $x_{i_1}, x_{i_2},\ldots $ where the first letter of each element is $a$ (constant for all elements). Remove the first letter from each of those elements, getting the sequence $x_{i_1}', x_{i_2}',\ldots $. Then, the infinite sequence $$x_1, x_2, \ldots, x_{i_1-1}, x_{i_1}', x_{i_2}', x_{i_3}', \ldots$$ meets the criteria 1-2 and is "smaller" than $x_1, x_2, \ldots$, a contradiction.