Show a measure is 1

Let $O=\{ (x_1,x_2)\in \mathbb{R}^2 \,:\,0<x_1=x_2<1 \}$, $I=\{N\in\mathcal{B} \,:\, \lambda((0,1)\cap N)=0\}$, where $\lambda$ is the Lebesgue measure and $\mathcal{B}$ is the Borel sigma algebra.

Furthermore let $\mu(A) = \begin{cases} 0 & \exists A_1,A_2\in I,\,\text{ and } \exists B_1,B_2\in\mathcal{B},\, \text{ so } A\subseteq(A_1\times B_1)\cup (A_2\times B_2) \\ 1 & \text{otherwise} \end{cases} $

Show that $\mu(O)=1$.

My thought is a contradiction argument. If $\mu(O)=0$, then there must exist $N_1,N_2\in I$ and $B_1,B_2\in\mathcal{B}$ such that $O\subseteq (N_1\times B_1)\cup(N_2\times B_2)$. But since $O$ is a line from the point $(0,0)$ to the point $(1,1)$, there cannot be a set with Lebesguemeasure $0$ on the interval $(0,1)$ on either axis. There is therefore no way to "describe" the line, because it goes from $0$ to $1$ on both axis, and therefore there does not exist $N_1,N_2\in I$ so $O\subseteq (N_1\times B_1)\cup (N_2\times B_2)$

My problem is that I understand why is the measure $\mu(O)=1$, I just can't seem to find the right proof technique.

Your approach is fine. Assuming that $O\subseteq (N_1\times B_1)\cup (N_2\times B_2)$, we get $(0,1)\subseteq N_1\cup N_2$. But $\lambda((N_1\cup N_2)\cap (0,1))=0<\lambda((0,1))$, which gives the required contradiction.