Update: A bit of a digital paper chase led me, via David Robert's thesis (note that in the latest version, it is Chapter 5, section 2 that is most relevant), to this paper on the arXiv. The last sentence of the abstract is:
These hoop earring spaces provide a simple class of counterexamples to the claim that $\pi_{1}^{top}$ is a functor to the category of topological groups.
I recommend reading this article.
The proof given in the second paper is short enough that I think it reasonable to copy it out here. I shan't copy out the obvious diagram so need to establish some notation first:
- $m \colon \pi_1^{Top}(X,x) \times \pi_1^{Top}(X,x) \to \pi_1^{Top}(X,x)$ is the multiplication map in question
- $p \colon \operatorname{Hom}((S^1,1),(X,x)) \to \pi_1^{Top}(X,x)$ is the quotient map
- $\overline{m} \colon \operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x)) \to \operatorname{Hom}((S^1,1),(X,x))$ is the "upstairs" multiplication map. (It's a tilde in the original, but that isn't displaying correctly for me so I daren't use it.)
The proof then proceeds:
To show that $m$ is continuous, it suffices to show that $\overline{m}$ is continuous, for then if $U \subset \pi_1^{Top}(X,x)$ is open, $(p \times p)^{-1} m^{-1}(U) = \overline{m}^{-1}p^{-1}(U)$ is open, but by the definition of a quotient map, $(p \times p)^{-1} m^{-1}(U)$ is open if and only if $m^{-1}(U)$ is.
There then follows a proof that $\overline{m}$ is continuous, a fact that I trust does not need proving.
Comments on your comments:
We don't need the quotient map to be open since we are only ever dealing with preimage sets. It is certainly not always true that if $q \colon X \to Y$ is a quotient that $q(U)$ is open in $Y$ for every open $U$ in $X$. But it is true by definition that $q^{-1}(U)$ is open in $X$ if and only if $U$ is open in $Y$. This is because the topology on $Y$ is precisely that to make this true. So since we are only dealing with sets of the form $(p \times p)^{-1}(A)$ then the assertion is valid assuming that $p \times p$ is a quotient map.
Here, I find myself worried. A quick back-of-envelope check seems to show that one can't simply assume that the product of quotients is again a quotient in Top (a counterexample eludes me as I don't have Counterexamples in Topology to hand and I'm too used to dealing with "nice" spaces). It may be the case that for Hom-spaces then there's some magic that can be done (though such is not mentioned in the paper); but again the best that I can do on the back of an envelope is observe that (modulo some basepoint mess) by construction $\operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x))$ quotients to $\pi_1^{Top}((X,x) \times (X,x))$. But to proceed, one would need to know that $\pi_1^{Top}$ was a product-preserving functor. This is morally the same as saying that it is representable - which looks good since we have an obvious representing object $S^1$! However, this can't be made into a proper argument since although we have a representing object, we don't have an enriched Hom-functor $hTop \times hTop \to Top$ which to evaluate at $S^1$.
So I would look for a counterexample to the product of quotients being a quotient, and see where that leads you. Either you'll find a proper counterexample to the proposition in question, or you'll see why in this special case, such a counterexample could not occur.
(Of course, I may well be missing something obvious!)