If $f:X\to Y$ is an identification map and $K$ is a locally compact Hausdorff space then $f\times 1:X\times K\to Y\times K$ is an identification map. I am trying to understand the proof of Proposition 13.9 [p. 43] in G. Bredon's book Topology and Geometry. Bredon's proof of theorem 13.9 relies on two previous results, which I will list now for convenience. 
Proposition 8.2. If $X$ is compact then the projection $\pi_Y : X\times Y\to Y$ is closed.
Proposition 13.5. A surjection $f:X\to Y$ is an identification map $\iff$ (for all functions $g:Y\to Z$, ($gf$ is continuous $\iff$ $g$ is continuous).)
And here is Proposition 13.9.
Proposition 13.9. If $f:X\to Y$ is an identification map and $K$ is a locally compact Hausdorff space then $f\times 1:X\times K\to Y\times K$ is an identification map. 
Proof. Suppose that $g:Y\times K\to W$ and let $h=g\circ (f\times 1): X\times K\to W$. Then, by Proposition 13.5, it suffices to prove that $h$ is continuous $\Rightarrow$ $g$ continuous. Let $U\subset W$ be open and suppose that $g(y_0,k_0)\in U$. Let $f(x_0)=y_0$. Then $h(x_0,k_0)=g(y_0,k_0)\in U$. Therefore there is a compact neighborhood $N$ of $k_0$ such that $h(x_0\times N)\subset U$. Put $A=\{y\in Y: g(y\times N)\subset U\}$. Then $y_0\in A$ and it suffices to show that $A$ is open. Thus it suffices to show that $f^{-1}(A)$ is open. Now
$$f^{-1}(A)=\{x\in X: h(x\times N)=g(f(x)\times N)\subset U\}$$
and so $X-f^{-1}(A)=\pi_X(h^{-1}(W-U)\cap (X\times N))$ is closed by Proposition 8.2. $\square$
Now, I understand Bredon's proof up until the point where he claims that: "Therefore there is a compact neighborhood $N$ of $k_0$ such that $h(x_0\times N)\subset U$." Since $K$ is locally compact, I know that there must be a compact neighborhood of $k_0$, but how does he deduce that we can find a compact neighborhood $N$ such that $h(x_0\times N)\subset U$? 
He then says: "Put $A=\{y\in Y: g(y\times N)\subset U\}$. Then $y_0\in A$ and it suffices to show that $A$ is open. Thus it suffices to show that $f^{-1}(A)$ is open." As I said above, I'm not really sure how he finds his compact neighborhood $N$ of $k_0$, so I do not see exactly where his set $A$ comes from and why it is useful to prove that $A$ is open. Our goal is to show that $g$ is continuous, so I thought we were trying to show that $g^{-1}(U)$ is open in $Y\times K$ and I do not see how showing that $A$ is open shows that $g^{-1}(U)$ is open. This is my main difficulty; that is, trying to understand why showing that $A$ is open proves that $g$ is continuous.
I think I understand the rest of the proof.
Thanks in advance.
 A: 
Since $K$ is locally compact, I know that there must be a compact neighborhood of $k_0$, but how does he deduce that we can find a compact neighborhood $N$ such that $h(x_0\times N)\subset U$?

$K$ is by assumption Hausdorff, so if $k_0$ has a compact neighbourhood $C$, then the compact neighbourhoods of $k_0$ form a neighbourhood base at $k_0$.
For if $V$ is any neighbourhood of $k_0$, then $U := \overset{\Large \circ}{V}\cap \overset{\Large\circ}{C}$ is an open neighbourhood of $k_0$ contained in $C$, and by the Hausdorff property and the compactness of $C\setminus U$, there is an open neighbourhood $W$ of $k_0$ and an open set $O$ containing $C\setminus U$ with $W\cap O = \varnothing.$ Then $\overline{W\cap U} \cap O = \varnothing$, and $\overline{W\cap U}$ is a neighbourhood of $k_0$ (since $W$ and $U$ are), and it is contained in $C$ (since $U$ and hence $\overline{U}$ is) and closed, hence compact. But since
$$\overline{W\cap U} \subset \overline{W}\cap\overline{U} \subset (K\setminus O) \cap C = C\setminus O \subset C\setminus (C\setminus U) = U \subset V,$$
we see that $V$ contains a compact neighbourhood of $k_0$. Since $V$ was an arbitrary neighbourhood of $k_0$, the assertion that $k_0$ has a neighbourhood base of compact neighbourhoods follows.
Note that only the existence of one compact neighbourhood of $k_0$ and the Hausdorffness of $K$ are needed for the conclusion, not that any other point also has a compact neighbourhood.
So for the conclusion that $k_0$ has a compact neighbourhood $N$ such that $h(x_0\times N) \subset U$, all we need is that $k_0$ has any neighbourhood $N_0$ such that $h(x_0\times N_0) \subset U$. But the assumed continuity of $h$ immediately yields that.

Our goal is to show that $g$ is continuous, so I thought we were trying to show that $g^{−1}(U)$ is open in $Y\times K$ and I do not see how showing that $A$ is open shows that $g^{−1}(U)$ is open.

Bredon shows the continuity of $g$ not by directly showing that $g^{-1}(U)$ is open for open $U\subset Z$, but by showing that $g$ is continuous at every point $(y_0,k_0) \in Y\times K$. So the task is to find, for every (open) neighbourhood $U$ of $g(y_0,k_0)$, a neighbourhood $W$ of $(y_0,k_0)$ such that $g(W) \subset U$. That is achieved by exposing a neighbourhood $W = V\times N$ with that property. As the first step, we saw above that there is a compact neighbourhood $N$ of $k_0$ such that $g(y_0\times N) \subset U$. By the definition of $A$, we have $g(A\times N) \subset U$. Now if one shows that $A$ is a neighbourhood of $y_0$, the proof is complete. Since $y_0 \in A$, showing that $A$ is open suffices. And since $f$ is an identification, $A$ is open if and only if $f^{-1}(A)$ is open.
