For $W=\cup_{U\in\mathcal U} U$ show that there exists $U_1,\dots,U_n: \ \sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W)$

I don't think the exact same argument works here but is there a similar approach to get at least pairwise disjoint sets in the case of this exercise? And further such that (1) holds?

The proposed argument is used to prove the Vitali covering lemma. As Martin Argerami noted, it needs that $C=x-C$ for some $x\in\Bbb R^d$. Then we can replace it by a set $C’=C-x/2=-C'$ containing the origin of $\Bbb R^d$ and either (provided $C’$ is non-empty) construct Minkowski functional to endow $\Bbb R^d$ with the metric allowing to apply Vitali covering lemma for a metric space or directly providing the required claim: if $$x_i+r_iC’\cap x_j+r_jC’\ne\varnothing$$ and $r_i\ge r_j$ then $ x_j+r_jC’ \subset x_i+3r_iC'$. Indeed, let $$x\in x_i+r_iC’\cap x_j+r_jC’$$ be an arbitrary point and $y\in x_j+r_jC’$. Then $$y\in x_j + r_jC' \subset x-r_jC’+ r_jC’\subset x_i+r_iC’-r_jC’+ r_jC’\subset x_i+3r_iC’.$$

But the condition $C=x-C$ was missed and the argument doesn’t work now.

Moreover, the exercise claim is wrong as the following example shows. Indeed, consider a simplex

$$C=C_d=\{(x_1,\dots,x_d): x_i>0\mbox{ for each }i\mbox{ and } x_1+\dots+x_d<1\}.$$

We have $\lambda(C_d)=\frac 1{d!}$. Let $\mathcal U=\{x+C: x\in -C\}$. Then $W=\bigcup U=C-C$. Since each member of $\mathcal U$ contains the origin of $\Bbb R^d$, $\mathcal U$ has no disjoint subsets.

Let’s calculate $\lambda(W)$. For each $\delta=(\delta_1,\dots,\delta_d)\in \{-1,1\}^d$ put

$$W_\delta=\{(x_1,\dots, x_d)\in W: \forall i (\delta_ix_i>0) \}.$$

Let $\delta_+=\{1\le i\le d: \delta_i=1\}$, $\delta_-=\{1\le i\le d: \delta_i=1\}$, and $\pi_+$ and $\pi_-$ be the projections of the product $\Bbb R^n=\Bbb R^{\delta_+}\times \Bbb R^{\delta_-}$ into its factors $\Bbb R^{\delta_+}$ and $\Bbb R^{\delta_-}$, respectively. Put $k=|\delta_+|$. It is easy to see that if $1\le k\le d-1$ then $\pi_+(W_\delta)$ is a natural copy of $C_k$, $\pi_-(W_\delta)$ is a natural copy of $C_{d-k}$, and $W_\delta=\pi_+(W_\delta)\times \pi_-( W_\delta)$. Thus $\lambda(W_\delta)=\frac 1{k!} \tfrac 1{(d-k)!}$.

So we have $$\lambda(W)=\sum\{\lambda(W_\delta): \delta\in \{-1,1\}^d \} =$$ $$\sum\left\{\frac 1{k!} \frac 1{(d-k)!}: \delta\in \{-1,1\}^d \mbox{ and } |\delta_+|=k\right\}=$$ $$\sum_{k=0}^d \frac 1{k!} \frac 1{(d-k)!}\cdot |\{\delta\in \{-1,1\}^d \mbox{ and } |\delta_+|=k\}|=$$ $$\sum_{k=0}^d\frac 1{k!} \frac 1{(d-k)!}{d\choose k}=\frac 1{d!}\sum_{k=0}^d {d\choose k}^2=\frac 1{d!} {2d\choose d}=\lambda(C){2d\choose d}$$ (see here) for the last equality.

Finally, Robbins’ bounds imply that

$$\frac {4^{d}}{\sqrt{\pi d}}\exp\left(-\frac {1}{8d-1}\right)<{2d\choose d}<\frac {4^{d}}{\sqrt{\pi d}}\exp\left(-\frac {1}{8d+1}\right).$$