Let me clarify the question first.
- $\mu$ is a probability distribution over the sample space $S=\{0,1\}^n \times \{0,1\}^n$.
- A combinatorial rectangle (or just rectangle) $R$ is a subset of $S$ of the form $A\times B$, where $A\subseteq \{0,1\}^n$ and $B\subseteq \{0,1\}^n$.
- For all $x\in \{0,1\}^n$, we can think of $x$ as the characteristic vector of a subset of $\{1, \cdots, n\}$. That is, $x=(x_1, x_2, \cdots, x_n)$ where $x_i=1$ if and only if $i$ is in that subset. We may abuse $x$ to mean that subset.
- We define $Disj(x, y) = 1$ for $x,y\in \{0,1\}^n$ if and only if the subset $x$ and the subset $y$ are disjoint. ($Disj(x, y) = 0$ if and only if they are not disjoint.)
For the sake of contradiction, suppose that $Disc_\mu(Disj) < \frac{1}{2n+1}$.
So, $Disc_\mu(S,Disj) = \left|\Pr_\mu [Disj(x,y)=1] - \Pr_\mu[Disj(x,y) =0] \right| < \frac{1}{2n+1}\tag{1},$
and $Disc_\mu(R_i,Disj) \lt \frac{1}{2n+1}\tag{2},$
where $R_i=X_i\times X_i$, $X_i$ is the set of all $x\in \{0,1\}^n$ that are subsets of $\{1, \cdots, n\}$ that contain $i$.
Since $Disj(x,y)=1$ means $x,y\in X_i$ for some $i$,
$$\begin{aligned}\Pr_\mu [Disj(x,y)=1] &\le\sum_{i=1}^n\Pr_\mu [(x,y)\in X_i\times X_i]\\
&=\sum_{i=1}^n\left|\Pr_\mu[(x,y) \in R_i \wedge Disj(x,y)=1] - \Pr_\mu[(x,y) \in R_i \wedge Disj(x,y) =0] \right|\\
&=\sum_{i=1}^n Disc_\mu(R_i,Disj) \lt \frac{n}{2n+1}
\end{aligned}$$
Combining the inequality (1), we know that $$\Pr_\mu [Disj(x,y)=0] \lt \frac {n+1}{2n+1}.$$
So $\Pr_\mu [Disj(x,y)=1] + \Pr_\mu[Disj(x,y) =0] <1$. However, the left hand side should be 1 since $Disj(x,y)$ is either 1 or 0. That is a contradiction. QED.