@TimCampion You get a get a bijection on underlying sets even if you are working in the category of all topological spaces. k-ification turns $|X \times Y| \to |X| \times |Y|$ into a homeomorphism, but note that k-ification of a space $A$ does not change which maps $S^k \to A$ are continuous, so we obtain a bijection $\mathbf{Top}(S^k,|X \times Y|) \to \mathbf{Top}(S^k,|X| \times |Y|)$, which is enough to detect weak homotopy equivalences.

So, to be clear, it is the bijection $\mathbf{Top}(S^k,|X \times Y|) \to \mathbf{Top}(S^k,|X| \times |Y|)$ that I think a pre-convenient-category-topologist could get their hands on.

@ArunDebray Thanks, nice! When we have a particular output which is identified with some other construction, the (variant of) Eilenberg-Steenrod axiom allows us to perform actual computation. I thought it's not computationally useful because a few examples I know about identifying functor with factorization homology is to prove ES axioms, but maybe here identification is given otherwise?

@AdrianClough And why is it the case that k-ification turns the comparison map into an iso?

I think the point is that a map from a compact space can only meet the interiors of finitely many cells and so factors through the geometric realization of a finite subcomplex.

Or am I missing something?

@EdoardoLanari I mean, there are arguments like the one you find on the nlab, but the argument there explicitly uses that your convenient category of spaces is cartesian closed. So I regard this argument as using the convenient category in an essential way. I'm in the dark as to how you'd do this if you didn't know about the convenient category.

@ReidBarton This sounds promising... the issue is that $|X| \times |Y|$ is not itself a CW complex. But I guess you could observe that each projection $|X \times Y | \to |X| \times |Y| \to |X|$ and $|X \times Y| \to |X| \times |Y| \to |Y|$ factors through a finite subcomplex $|X'|, |Y'|$, so that the map factors through $|X'| \times |Y'|$...

Then since $|X'| \times |Y'| = |X' \times Y'| \subseteq |X \times Y|$, you've lifted your homotopy group element along the map. Ok!