Recall that the standard middle-third Cantor set $C$ has lower box/Minkowski dimension $d=\log 2/\log 3$ (same as upper, same as Hausdorff). One can reduce this dimension by changing the construction so that at certain steps we throw out a lot more than middle-third. Let's say we remove middle $\lambda_k/3$ portion instead of $1/3$ at step $n=n_k$. The sequence $\{n_k\}$ will grow super-exponentially, for example $n_k = 2^{2^k}$.
The resulting set, call it $A$, can be covered by $N = 2^{n_k}$ intervals of length
$$\epsilon = 3^{-n_k}\prod_{j=1}^k \lambda_j$$
We'll choose these so that
$\log N /\log(1/\epsilon) \to d/2$. So, we need
$$ \frac{n_k \log 2}{n_k\log 3 + \sum_{j=1}^k \log(1/\lambda_j)} \to \frac{d}{2}$$
which can be achieved by choosing $\lambda_k$ so that $\log(1/\lambda_k) = n_k \log 3$. So, $\lambda_k = 3^{-n_k}$. Note that the sum over $j$ is asymptotic to its largest term, due to super-exponential growth of $n_k$.
What happens if we try to cover this set by elements of a different scale? After $n_k+m$ steps, where $n_k+m < n_{k+1}$, there are $2^{n_k+m}$ intervals of length
$$\epsilon = 3^{-n_k-m}\prod_{j=1}^k \lambda_j$$
so
$$
\frac{\log N}{\log(1/\epsilon)} =
\frac{(n_k+m) \log 2}{(n_k+m)\log 3 + \sum_{j=1}^k \log(1/\lambda_j)}
\sim \frac{(n_k+m) \log 2}{(2n_k+m)\log 3}
$$
When $m\ge n_k$, this quotient is at least $2d/3$.
The upshot is that the lower box dimension of $A$ is $d/2$ but we can only achieve that by covering it at the scales between
$$
\epsilon_k = 3^{-n_k} \prod_{j=1}^k 3^{-n_j} \quad\text{and}\quad 3^{-n_k} \epsilon_k
$$
which is a relatively narrow range of scales, considering that $\epsilon_{k+1} = 3^{-n_{k+1}} \prod_{j=1}^{k+1} 3^{-n_j}$ is much smaller than $3^{-n_k} \epsilon_k$. Outside of this range of "convenient" scales, the covering numbers yield the dimension of at least $2d/3$.
It remains to repeat this construction but using $\tilde n_k=2n_k$ instead. The resulting set, call it $B$, will again have lower box dimension $d/2$. But the union $A\cup B$ has lower box dimension at least $2d/3$, because the "convenient" scales for $A$ are not "convenient" for $B$.
