Being measurable w.r.t. $m\times\nu$ doesn't make sense, and furthermore you don't even use the product measure in this exercise.
Instead, you should have specified which sigma-algebra you equip $[0,1]$ with - both when you're speaking of $m$ and when you're speaking of $\nu$. You could for example consider the measure-space $([0,1],\mathcal{E},m)$ and $([0,1],\mathcal{F},\nu)$, where $\mathcal{E}=\mathcal{B}([0,1])$ is the Borel sigma-algebra on $[0,1]$ and $\mathcal{F}$ could be $\mathcal{B}([0,1])$ or even the power set $\mathcal{P}([0,1])$. But let us assume that $\mathcal{F}=\mathcal{B}(\mathbb{R})$ since this is the smallest of the two.
Now you should show that $$D\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}^2),$$
i.e. that $D$ belongs to the product-sigma-algebra of $[0,1]\times [0,1]$. One strategy for that is to show that $D$ is closed in $\mathbb{R}^2$. This ensures that the sections $D_x=\{y\in\mathbb{R}\mid (x,y)\in D\}$ and $D_y=\{x\in\mathbb{R}\mid (x,y)\in D\}$ belongs to $\mathcal{B}(\mathbb{R})$.
For (2) you just evaluate the inner integrals first: For a fixed $y\in [0,1]$ we have that $\chi_D(x,y)=\chi_{D_y}(x)=1$ if and only if $x=y$ and zero otherwise. Therefore,
$$
\int_{[0,1]}\chi_D(x,y)\,m(\mathrm dx)=\int_{[0,1]}\chi_{D_{y}}(x)\,m(\mathrm dx)=m(\{y\})=0,
$$
for all $y\in [0,1]$.
For the right-hand side we have that for a fixed $x\in [0,1]$:
$$
\int_{[0,1]}\chi_D(x,y)\,\nu(\mathrm dy)=\int_{[0,1]}\chi_{D_{x}}(y)\,\nu(\mathrm dy)=\nu(\{x\})=1.
$$
Is this a contradiction to Tonelli/Fubini's theorem? (This is probably the key point of the exercise).
Recall that simple functions are just finite linear combinations of characteristic functions of measurable sets. In this context our measure space is discrete and all sets are measurable, so a simple function is just a linear combination of (characteristic functions of) points. More precisely, in our context every simple function takes the form
$$s(n)=\sum_{k=1}^m s_k \mathbf{1}_{\{a_k\}}(n),$$
for some $m\in\mathbb{N}\cup\{\infty\}$ and $(a_k)_{k=1}^m\subset \mathbb{N}$ pairwise different numbers. Note how $m=\infty$ is also allowed, because also infinite sets of natural numbers are measurable.
Say that $s$ is one of the objects that the supremum is taken over, so a simple function with $0\le s(n)\le f(n)$ for all $n$. In particular $s_k=s(a_k)\le f(a_k)$.
Then
$$\int_{\mathbb{N}} s\,d\mu = \sum_{k=1}^m s_k\le \sum_{k=1}^m f(a_k)\le \sum_{k=1}^\infty f(k).$$
In the last step we used that $f$ is non-negative. By definition of the supremum as the least upper bound this implies
$$\int_{\mathbb{N}} f\,d\mu \le \sum_{k=1}^\infty f(k).$$
Best Answer
Counting measure is just summation!
To see this, you can approach from a few different angles; how about we consider the Monotone Convergence Theorem. To that end, for $n\in\mathbb{N}$, define $f_n:\mathbb{N}\to\mathbb{R}$ by $$ f_n(k)=\begin{cases}f(k) & \text{if }1\leq k\leq n\\ 0 & \text{else}\end{cases}. $$ Then clearly, as $n\to\infty$, $f_n\to f$ pointwise; it is also monotone increasing, because $f(k)\geq0$ for all $k\in\mathbb{N}$. So, by the MCT, $$ \int f_n\,d\mu\to\int f\,d\mu\text{ as }n\to\infty. $$ Now, consider these $f_n$. Note that we can write $$ \mathbb{N}=\{1\}\cup\{2\}\cup\cdots\cup\{n\}\cup\{n+1,n+2,\ldots\}, $$ and that these sets are all measurable. So, $$ \begin{align*} \int f_n\,d\mu&=\int_{\{1\}}f_n\,d\mu+\cdots+\int_{\{n\}}f_n\,d\mu+\int_{\{n+1,n+2,\ldots\}}f_n\,d\mu\\ &=\int_{\{1\}}f_n(1)\,d\mu+\cdots+\int_{\{n\}}f_n(n)\,d\mu+\int_{\{n+1,n+2,\ldots\}}0\,d\mu, \end{align*} $$ where we have used that $f_n$ is constant on each of these sets, by definition. So, we see that $$ \int f_n\,d\mu=1\cdot f_n(1)+1\cdot f_n(2)+\cdots+1\cdot f_n(n)+0=f(1)+\cdots+f(n). $$ So, we have that $$ \int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu=\lim_{n\to\infty}(f(1)+\cdots+f(n))=\sum_{k=1}^{\infty}f(k). $$ The boundedness is of no consequence here -- since our terms are non-negative, the series either converges or diverges to $\infty$; in either case, the integral is exactly the sum.