It is better if you solve this problem on your own. The solution you quote as too confusing and inconsistent with notation.
As you already found out:
- a. In the topological space $X=(\mathbb{R}\times\mathbb{R},\tau_d\otimes\tau_e)$, where $\tau_d$ is the discrete topology on $\mathbb{R}$ and $\tau_e$ is the usual Euclidean topology, a compact set $K$ is of the form $\bigcup^n_{j=1}\{x_j\}\times K_j$, where $\{x_1,\ldots,x_n\}\subset\mathbb{R}$, and the $K_1,\ldots,K_n$ are compact sets in $(\mathbb{R},\tau_e)$. It is also clear that $X$ is Hausdorff and locally compact.
- b. If $f\in\mathcal{C}_c(X)$, then there are points $\{x_j:1\leq j\leq m\}$ and a sequence of functions $\phi_j\in\mathcal{C}_c((\mathbb{R},\tau_e))$, $1\leq j\leq m$ such that
$$ f(x_j,y)=\phi_j(y)\qquad 1\leq j\leq m, \quad y\in\mathbb{R}$$
The map $\Lambda(f):=\sum^m_{j=1}\int_\mathbb{R} \phi_j(y)\,dy$ clearly defines a linear and positive functional on $\mathcal{C}_c(X)$ (why?)
- c. Form the Riesz representation theorem there is a measure $\mu$ on a $\sigma$-algebra $\mathfrak{M}_\Lambda$ that contains $\mathscr{B}(X)$ such that
$$\Lambda f=\int_X f\,d\mu,\qquad f\in\mathcal{C}_c(X)$$
To determine the marginals of $\mu$ over $(\mathbb{R},\tau_d)$ and $(\mathbb{R},\tau_e)$ first notice that for any fixed $\phi\in\mathcal{C}_c((\mathbb{R},\tau_e))$, and any $x_1,x_2\in\mathbb{R}$,
if $f_i(x,y)=\mathbb{1}_{\{x_i\}}(x)\phi(y)$, $i=1,2$, then $f_i\in\mathcal{C}_c(X)$ and
$$\Lambda f_1=\int_{\mathbb{R}}\phi(y)\,dy=\Lambda f_2$$
The Riesz representation theorem applied to $(\mathbb{R},\tau_e)$ and $\tilde{\Lambda}:\phi\mapsto \int_{\mathbb{R}}\phi(y)\,dy$, $\phi\in\mathcal{C}_c((\mathbb{R},\tau_e))$ implies that for any compact set $B\subset\mathbb{R}$, there is a sequence $\phi_n\in \mathcal{C}_c((\mathbb{R},\tau_e))$ such that $\phi_n\searrow\mathbb{1}_B$ point wise, and
$$\int_\mathbb{R}|\mathbb{1}_B(y)-\phi(y)|\,dy\xrightarrow{n\rightarrow\infty}0$$
Define $g^i_n(x,y)=\mathbb{1}_{\{x_i\}}(x)\phi_n(y)$, $ i = 1,2$. Then each $g^i_n\in\mathcal{C}_c(X)$ $ i = 1,2$, and $n\in\mathbb{N}$. Furthermore
$$
\int_X|\mathbb{1}_{\{x_i\}}(x)\mathbb{1}_B(y)-g^i_n(x,y)|\,\mu(dx,dy)=\int_{\mathbb{R}}|\mathbb{1}_B(y)-\phi_n(y)|\,dy\xrightarrow{n\rightarrow\infty}0$$
and so,
$$\int_X\mathbb{1}_{\{x_1\}}\mathbb{1}_{B}(y)\,\mu(dx,dy)=\int_X\mathbb{1}_{\{x_2\}}\mathbb{1}_{B}(y)\,\mu(dx,dy)=\lim_n\int_{\mathbb{R}}\phi_n(y)\,dy=\lambda(B)$$
That is,
$$\begin{align}\mu(\{x\}\times B)=\lambda(B)\tag{1}\label{one}\end{align}$$
for all $x\in\mathbb{R}$ and any compact set $B$.
(Solution to 3.) If $K\subset\mathbb{R}\times\{0\}=E$ is compact, then $K=\{x_1,\ldots,x_m\}\times\{0\}$ for some points $x_j\in\mathbb{R}$ and so $\mu(K)=\sum^m_{j=1}\mu(\{x_j\}\times\{0\})=0$ by \eqref{one}.
(Solution to 4.) The Riesz representation also implies that for any measurable $A\subset X$
$$ \mu(A)=\inf\{\mu(G):G\,\text{open,}\, A\subset G\}$$
Any open set $G$ that contains $E$, also contains sets of the form $U=\bigcup_{x\in\mathbb{R}}\{x\}\times(-a_x,a_x)$, $a_x>0$. Since $\mathbb{R}$ is uncountable, there is $n_0\in\mathbb{N}$ such that $X_{n_0}=\{x:a_x>\frac{1}{n_0}\}$ is uncountable. Thus, for such $G$
$$\mu(G)\geq\mu^*(U)\geq\sup_{J\subset X_{n_0}}\sum_{x\in J}2a_x=\infty,$$ where the sup is over all finite subsets of $X_{n_0}$. That is $\mu^*(E)=\infty$.
Best Answer
If we know $\Lambda f = \int_X fd\mu$ for some $\mu$, then if $K$ is compact we can find some function $g$ that is $1$ on $K$ and compactly supported on $X$ and then this integral is at least $\mu(K)$ and must also be finite, as $\Lambda(f)$ is. So $\mu$ is then always be finite on compact sets. So for $(a)$ to hold at all, $(b)$ must be satisfied for the $\mu$ we want. I think that's all he's saying.