Hausdorff dimension of $A \times A$ (Folland's exercise 11.2.15) The next question is regarding Hausdorff measure, it was taken from "Real Analysis" by Folland (question 11.15).

If $A \subset \mathbb{R}^n$ has Hausdorff dimension of $p$, then $A \times A \subset \mathbb{R}^{2n}$ has Hausdorff dimension $\ge 2p$.

My attempt:
We can see that for any $B \subset \mathbb{R}^n$, $\operatorname{diam}(B \times B) \le \sqrt{2\operatorname{diam}(B)}$, and if we assume that $H_p(A)< \infty$, then for any $\delta > 0$,
We can find $\{B_i\}_{i=1}^{\infty} \subset R^n, \operatorname{diam}(B_i) < \delta , A \subset \bigcup_{i=1}^{\infty}{B_i}, s.t$
$$\sum_{i=1}^{\infty}{\operatorname{diam}(B_i)^p} \le H_p(A) < \infty$$
So by choosing $\{ B_i \times B_i \}_{i=1}^{\infty}$ we will get that $\operatorname{diam}(B_i \times B_i) \le \sqrt{2\delta}, A \times A \subset \bigcup_{i=1}^{\infty}{B_i \times B_i}$
$$ \sum_{i=1}^{\infty}{\operatorname{diam}(B_i \times B_i)^{2p}} \le \sum_{i=1}^{\infty}{\sqrt{2\operatorname{diam}(B_i)}^{2p}} = 2\sum_{i=1}^{\infty}{\operatorname{diam}(B_i)^p} \le 2H_p(A) < \infty
$$
I believe that from this we can conclude that the dimension of $A \times A \ge 2p$ but I'm not sure why, and there is still the case where $H_p(A) = \infty$.
Thanks.
 A: $\newcommand{\H}{\mathscr{H}}\newcommand{\diam}{\operatorname{diam}}$EDIT: I had to temporarily delete this answer because your attempt had mislead me. We were both wrong! So, I no longer have a suggested solution for you. But I'll leave this here for now as it informs you about what you still need to do / have done wrong.

Your writings were a bit unclear for me, so let me write up what you have attempted to show:

If $A$ has Hausdorff dimension $p$ and $\H_p(A)<\infty$ then for all $\delta>0$ the quantity $\H^{\delta}_{2p}(A\times A)$ is finite.

That implies: $$\H_{2p}(A\times A):=\sup_{\delta>0}\H_{2p}^\delta$$Is finite. That actually only implies the Hausdorff dimension of $A\times A$ is no more than $2p$, whereas you actually want to show the reverse inequality. Because, if the dimension is $d\in[0,\infty]$, it is known that: $$\H_\alpha(A\times A)=\begin{cases}\infty&\alpha<d\\0&\alpha>d\end{cases}$$If $d>2p$ then we have a contradiction to the observation that $\H_{2p}(A\times A)<\infty$. However, it could still be the case that $\H_{2p}(A\times A)=0$, and $d$ might be less than $2p$. This would not contradict anything you have shown so far. However, your method of proof for the above was suspect.

What went wrong? Well, it's not actually true that the $B_i\times B_i$ cover $A\times A$. It is true that: $$\bigcup_{i\ge1}\bigcup_{j\ge1}B_i\times B_j$$Is a cover of $A\times A$ however. Do you see the difference?
Furthermore: $$\diam(B\times B)=\diam(B)\sqrt{2}$$With exact equality. Your inequality is not right if $\diam(B)>1$ (consider $B=[-1,1]$ and $B\times B=[-1,1]^2$ - the diameters are $2,2\sqrt{2}$ respectively and $2\sqrt{2}$ is not less than or equal to $\sqrt{2\cdot(2)}$).

Possible ways to continue:
To show the dimension of $A\times A$ is at least $2p$ it is equivalent to show that $\H_{d}(A\times A)=\infty$ whenever $d<2p$.
