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).
I am assuming that the product measure is the one induced by the product outer measure. (This avoids any issue with ambiguity of definition.)
Any set of the form $B \times A$, with $B$ Borel is measurable ($A$ is arbitrary).
Take the sets $S_n = \{(1,1)\} \cup\left( \cup_{k=0}^{n-1} [\frac{k}{n},\frac{k+1}{n})^2 \right)$. Clearly each $S_n$ is measurable, and hence $D = \cap_{n \ge 0} S_n $ is measurable.
We must have $(m \times \mu) D = \infty$.
To see this, suppose
$D \subset \bigcup_{n \ge 0} B_n \times A_n$, where $B_n$ is Borel, and $A_n$ is arbitrary. Let $D_n = D \cap (B_n \times A_n)$, and let $\pi_x((x,y)) = x$ and similarly $\pi_y((x,y)) = y$. Since $[0,1] \subset \cup_n \pi_x D_n$, we must have $m^* (\pi_x D_{n'} ) >0$ for some $n'$, where $m^*$ is the Lebesgue outer measure (using $m^*$ avoids having to worry about the measurability of $\pi_x D_{n'}$).
Hence $m B_{n'} \ge m^* (\pi_x D_{n'} ) > 0$. In addition, $\pi_x D_{n'}$ must be uncountable (otherwise $m^* (\pi_x D_{n'} ) $ would be zero). Furthermore, $\pi_y D_{n'} = \pi_x D_{n'}$, $\pi_y D_{n'} \subset A_{n'}$, hence $A_{n'}$ is also uncountable. Hence $m(B_{n'}) \mu(A_{n'}) = \infty$, and so $\sum_{n \ge 0} m(B_{n}) \mu(A_{n}) = \infty$ for any cover of $D$ by measurable rectangles. Hence $(m \times \mu) D = (m \times \mu)^* D =\infty$.
Best Answer
The left-sided integral is $m([0,1])=1$. Remember that $\mu$ is not $\sigma$-finite on $[0,1]$, so Fubini Theorem fails in this case.