Higman's lemma and a manuscript of Erdős and Rado

I didn't have much time when I wrote my initial answer, so here's an update.

It occurred to me that I ought to recommend Kruskal's classic paper "The theory of well-quasi-ordering: a frequently discovered concept" (JCTA (1972), 297–305), so I went to see if he had anything to say about the equivalence of (a) and (b). He doesn't seem to have anything to say about that, but he does have something to say about the reference you mention. On page 300, Kruskal writes that

Incidentally, Higman refers to an unpublished manuscript of Erdős and Rado which was probably an early version of [29] or of [4].

Kruskal's [4] is Erdős and Rado's solution to problem 4358 in the Monthly in 1952 (pp. 255–257).

Kruskal's [29] is Rado's paper "Partial well-ordering of sets of vectors" (Mathematika (1954), 89–95).

Another paper that Higman might have been referring to is Erdős and Rado's 1959 paper "A theorem on partial well-ordering of sets of vectors" (J. London Math. Soc. (1959), 222–224).

Anyway, that's only the first part of your question. The second part asks where to find the proof. Perspectives have changed since Higman, and condition (b) of your question is now often taken as the definition of a well-quasi-order (every infinite sequence has a good pair). To derive (a) from this definition, one typically appeals to Ramsey's theorem as Andreas Blass has done in his answer here. You can also find this proof in almost any text which includes an introduction to well-quasi-order. To mention one proof that is particularly succinct, you might look at the chapter of Diestel's Graph Theory that covers minors.

This is an answer not for the original question but for how to get the equivalence of (a) and (b) from the infinite Ramsey theorem (as requested in a comment). The implication from (a) to (b) is trivial. For the converse, I'm going to assume that "sequence" means one-to-one sequence (i.e., no repetitions), because otherwise a one-element set $A$ is a counterexample (because of the "strictly" in (a) and the non-strict $\preceq$ in (b)).

So assume (b) and let $(a_i)_{i\in\mathbb N}$ be a sequence of distinct elements of $A$. Partition the set $[\mathbb N]^2$ of $2$-element subsets $\{i<j\}$ of $\mathbb N$ into two parts by putting $\{i<j\}$ into the first part if $a_i\prec a_j$ and into the second part otherwise. By Ramsey's theorem, there is an infinite subset $H$ of $\mathbb N$ all of whose $2$-element subsets are in the same part.

If they're all in the first part, that means that, for all $i<j$ in $H$, we have $a_i\prec a_j$. In other words, the subsequence $(a_i)_{i\in H}$ of our original sequence is increasing with respect to $\prec$, as required for (a).

If they're all in the second part, that means that, for all $i<j$ in $H$, we have $a_i\not\prec a_j$. But then the sequence $(a_i)_{i\in H}$ violates our assumption (b).

(The last step is where I need that the sequence is one-to-one, because I need $a_i\not\preceq a_j$ in order to violate (b). If one allows sequences that are not one-to-one, then the best I can do is to partition $[\mathbb N]^2$ into three parts, according to whether $a_i\prec a_j$, $a_i=a_j$ or $a_i\not\preceq a_j$. If $H$ is an infinite homogeneous set as given by Ramsey's theorem, then the result is that $(a_i)_{i\in H}$ either is increasing as required for (a), or is constant, or violates (b). So what can go wrong in the not one-to-one case is essentially just the possibility of a constant sequence, as I indicated above in my remark about a one-element set $A$.)