We have the functor $K:\bf{Hauss}\to \bf{CGHauss}$ which assigns to each Hausdorff space, its Kellyfication. Then, we can define a product $\square :\bf{CGHauss}\times \bf{CGHauss}\to \bf{CGHauss}$ by $X\square Y=K(X\times Y).$
The proof that $[Z\square Y,X]\cong [Z,X^Y]$ then proceeds in the usual way by showing that the evaluation $e:X^Y\square Y\to X$ is universal:
for $h:Z\square Y\to X$ one defines $k:Z\to X^Y$ by $kz(y)=h(z,y)$ and then shows that $z\mapsto kz$ is continuous. MacLane does this by asserting that it is enough to consider the compact-open topology on $Y^X.$ Why is this so? Why does it suffice to work in a weaker topology?