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:

 1. $m \colon \pi_1^{Top}(X,x) \times \pi_1^{Top}(X,x) \to \pi_1^{Top}(X,x)$ is the multiplication map in question
 2. $p \colon \operatorname{Hom}((S^1,1),(X,x)) \to \pi_1^{Top}(X,x)$ is the quotient map
 3. $\overline{m} \colon \operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x)) \to \operatorname{Hom}((S^1,1),(X,x))$ 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:

1. 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_.

2. 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!)