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).
Well, for each $y\in\mathbb{R}$,
$$
\int 1_A(x,y)\,d\lambda(x)=\int 1_{\{y\}}(x)\,d\lambda(x)=0.
$$
On the other hand, for each $x\in\mathbb{R}$,
$$
\int 1_A(x,y)\,d\mu(y)=\int 1_{\{x\}}(y)\,d\mu(y)=1.
$$
Best Answer
All three are wrong. The first one doesn't make sense. The second and third are both trying to use Tonelli's theorem, but Tonelli's theorem does not hold here since the counting measure is not sigma-finite. A correct way to proceed is to use the definition of the product measure as the inf of sum of measures of measurable rectangles that cover your set. Assume that we use the Borel sigma algebra on $\mathbb{R}$. You can show that $(\mu \times \nu)(\{(x, x) : x \in [0, 1]\}) = \infty$. The case of $\mu$ and $\nu$ on the Borel subsets of $[0, 1]$ is a classic example of when all three things that are supposed to be equal in Tonelli's theorem are unequal.