Let $f:[0,1]^2\rightarrow \{0,1\}$ given by $f(x,y)=\chi_{x=y}(x,y)$. First I want to show that $f$ is measurable and then that $$\int_{[0,1]}\int_{[0,1]}fd\mu d\lambda\neq \int_{[0,1]}\int_{[0,1]}fd\lambda d\mu,$$ where $\lambda$ is the lebesgue measure and $\mu$ the Counting measure.
I know that $f$ is measureable iff $\{x=y\}$ is measurable, so I tried to write it as a union or intersection of sets of the form $\pi^{-1}_i(A)$ with $A\in B([0.1])$ where $\pi_i:[0.1]^2 \rightarrow [0.1]$ is the projection.
So I chose for a given point $c \in \{(x,x)\in [0,1]^2\}$ $$\{c\}=\pi_1^{-1}({x=c})\cap \pi_2^{-1}({y=c})$$ since
$$\pi_1^{-1}(c)=\{c\}\times [0.1] \ \ \ \pi_2^{-1}(c)=[0.1] \times \{c\}.$$ However the union of these points would be uncountable I think.
Provided that $f$ is measurable I still would not really know how to compute these integrals. I know the definition $\int f d\mu=\sum_{n=1}^\infty \alpha_n\mu(A_n)$ In this case I would probably only need $\int f d\mu=\mu(\{x=y\})$ which should be $\infty$. Additionally I am not sure how to compute $\lambda(\{x=y\})$.
You were very close to the answer. The only thing that you are missing is that $\lambda(\{x=y\})=0$. Note that every square with center at $(y,y)$ and length $l$ contains $\{x=y\}$. Therefore, $\lambda(\{x=y\}) \leq l^2$. Since the inequality holds for every $l$, $\lambda(\{x=y\}) =0$. Now you can show that one order of integration yields $0$ and the other one yields $\infty$.
Also, in order to show that $L=\{(x,x): x \in [0,1]\}$ is measurable, define $A_{i,n}=\{\frac{i-1}{n} \leq x,y \leq \frac{i}{n}\}$. Note that $\cap_{n=1}^{\infty}\cup_{i=1}^{n}{A_{i,n}}=L$. You can see this by noting that each $A_{i,n}$ is a square and the they cover $L$. Since the area of the squares go to $0$, only $L$ remains after you take the intersection.