I am trying to show that: for bounded $E,F\subset\mathbb{R}$, that $\overline\dim_{B}(E\cup F)=\max\{\overline\dim_{B}(E),\overline\dim_{B}(F)\}$. So far I have:
Let $N_{\delta}(S)$ the smallest number of closed balls of radius $\delta$ which cover a set $S$. Clearly $E,F\subset E\cup F$, so $\max\{\overline\dim_{B}(E),\overline\dim_{B}(F)\}\leq\overline\dim_{B}(E\cup F)$. For any $a,b>0$, $a+b\leq\max\{a,b\}$, so $N_{\delta}(E\cup F)\leq N_{\delta}(E)+N_{\delta}(F)\leq2\max\{N_{\delta}(E),N_{\delta}(F)\}$.
I don't know where to go from there. I feel like I'm really close but can't quite make the last step.
I think this is correct, hopefully some really clever person can confirm it.
Solution: $N_{\delta}(S)$ the smallest number of closed balls of radius $\delta$ which cover a set $S$. Clearly $E,F\subset E\cup F$, so $\max\{\overline{\dim}_{B}(E),\overline{\dim}_{B}(F)\}\leq\overline{\dim}_{B}(E\cup F)$.
Now, $N_{\delta}(E\cup F)\leq N_{\delta}(E)+N_{\delta}(F)\leq2\max\{N_{\delta}(E),N_{\delta}(F)\}$ and so $$\frac{\log(N_{\delta}(E\cup F))}{-\log(\delta)}\leq\max\left\{\frac{\log(N_{\delta}(E))}{-\log(\delta)},\frac{\log(N_{\delta}(F))}{-\log(\delta)}\right\}+\frac{\log(2)}{-\log(\delta)}.$$ Letting $\delta\to0$ gives that $\overline{\dim}_{B}(E\cup F)\leq\max\{\overline{\dim}_{B}(E),\overline{\dim}_{B}(F)\}$.