In the proof pg 191, 5.9.10 we are trying to prove the exponential correspondence,
$$e_K:K(X \times_k Y, Z) \rightarrow K(X,K(Y,Z)). $$
Ronnie wrote
$e_K$ corresponds to the map $$e': K(X\times_k Y, Z) \times_k X \rightarrow K(Y,Z) $$ which corresponds to $$e'':K(X \times_k Y,Z) \times_k X \times _k Y \rightarrow Z$$ which is continuous. Hence $e'$ is continuous, hence $e_K$ is continuous.
Where did the first, second and third continuity come from?
I have a long winded proof: but it boils down to proving that
For $k$-spaces, $X,Y$, the evaluation map, $$e: K(X,Y) \times _k X \rightarrow Y, (f,x) \mapsto f(x) $$ is continuous.
Proposed proof: $e$ is continuous by 5.9.6, pg187 iff $e(s \times t)$ is so for all test maps $s:B \rightarrow K(X,Y)$, $t:C \rightarrow X$. This is continuous, by 5.9.5 since it is the restriction of $e(1 \times t)$.
Let us first understand the exponential map on the level of sets. For sets $A, B$ let $[A,B]$ denote the set of all functions $f : A \to B$. Define $e = e^{A,B,C}: [A \times B,C] \to [A,[B,C]]$ as follows: For $f : A \times B \to C$ and $a \in A$ define a function $f_a : B \to C, f_a(b) = f(a,b)$. Then $e(f)$ is defined by $e(f)(a) = f_a$. It is easy to verify that $e$ is a bijection of sets.
Now we come to topology. Given topological spaces $X, Y$, there are various ways to define a topological function set $\mathcal{T}(X,Y) \subset [X,Y]$. The simplest example is $Y^X$ = set of all continuous maps $X \to Y$. Ronnie Brown introduces $\mathcal{K}(X,Y) = Y^{kX}$. By construction $X^Y \subset \mathcal{K}(X,Y)$, and if $Y$ is a k-space, then $X^Y = \mathcal{K}(X,Y)$.
The big question is how to define a topology on each $\mathcal{T}(X,Y)$, i.e. to define a function space whose underlying set is $\mathcal{T}(X,Y)$. Its topological properties should of course be as nice as possible.
For $Y^X$ the standard approach is to use the compact-open topology (see any book on general topology). Ronnie Brown prefers to work with $\mathcal{K}(X,Y)$ and defines a variant of the compact-open topology. In general $Y^X$ is not a subspace of $\mathcal{K}(X,Y)$. Another space is introduced as $K(X,Y) = k\mathcal{K}(X,Y)$; its topology is in general finer than that of $\mathcal{K}(X,Y)$.
Similarly there are various ways to define a topology on the cartesian product $X \times Y$ of the sets $X,Y$, in that way producing a topological space $\mathcal{P}(X,Y)$. It is standard to understand $X \times Y$ as being endowed with the product topology. Ronnie Brown uses $X \times_k Y = k(X \times Y)$ whose topology is in general finer than that of $X \times Y$. We get in particular $\mathcal{K}(X \times Y,Z) = Z^{X \times_k Y}$.
In what follows we always understand $\mathcal{T}(X,Y)$ as a function space; its topology is given by a rule applicable to any pair of spaces.
We have $\mathcal{T}(\mathcal{P}(X,Y),Z) \subset [X \times Y,Z]$ and $\mathcal{T}(X, \mathcal{T}(Y,Z)) \subset [X, [Y,Z]]$. Concerning the exponential map $e$ we would like to have
(1) $e$ maps $\mathcal{T}(\mathcal{P}(X,Y),Z))$ bijectively onto $\mathcal{T}(X, \mathcal{T}(Y,Z))$.
(2) $e : \mathcal{T}(\mathcal{P}(X,Y),Z)) \to \mathcal{T}(X, \mathcal{T}(Y,Z))$ is a homeomorphism.
We now need a little lemma which is not explicitly stated in the book:
Let $S$ be any space, $X$ be a k-space and $f : X \to kS$ be a function. Then $f$ is continuous if and only if $i \circ f : X \to S$ is continuous, where $i :kS \to S$ is the identity of sets.
One half is trivial because $i$ is continous. So let $i \circ f$ be continuous. To show that $f$ is continuous it suffices to show that all $f \circ t$ are continuous, where $t : C \to X$ with compact Hausdorff $C$ (see the definition of a k-space). But the continuity of $f \circ t$ follows from the remark on top of p. 185.
This implies that for a k-space $X$ we have $\mathcal{K}(X, \mathcal{K}(Y,Z)) = \mathcal{K}(X, k\mathcal{K}(Y,Z)) = \mathcal{K}(X, K(Y,Z))$.
This explains the equality of sets claimed in the beginning of the proof of 5.9.10.
5.9.8 states that $e$ maps $\mathcal{K}(X \times Y,Z)$ bijectively onto $\mathcal{K}(X, \mathcal{K}(Y,Z))$. But we have $X \times_k Y = k(X \times Y)$ so that $\mathcal{K}(X \times_k Y,Z) = \mathcal{K}(X \times Y,Z)$ because $kkS =kS$ for any space $S$ (in particular for $S = X \times Y$). This explains why 5.9.8 shows that $e_K = e_K^{X,Y,Z}$ is a well-defined bijection.
Consider the exponential bijection $e^{U,V,W}$ for $U = K(X \times_k Y,Z), V = X, W = K(Y,Z)$. We have $e' =(e^{U,V,W})^{-1}(e_K)$. 5.9.8 says that $e_K$ is continuous if and only if $e'$ is. Similarly we see that $e'$ is continuous if and only the evaluation map $e''$ is. Using once more the exponential bijection for $U' = K(X \times_k Y,Z), V' = X \times_k Y, W' = Z$, we see that $e''$ is continuous if and only if $e^{U',V',W'}(e'')$ is. But $e^{U',V',W'}(e'')$ is the identity.
Nice application of the exponential map on various levels, isn't it? But of course complicated.
The usefulness of k-spaces in the context of function spaces was recognized by Norman E. Steenrod (see reference [Ste67] in the book).
https://projecteuclid.org/download/pdf_1/euclid.mmj/1028999711