Define
$K_{0}=[0,1] $
$K_{1}=[0,1/8]\cup[1-1/8,1] $
$K_{2}=[0,1/8^{2}]\cup[1/8-1/8^{2},1/8]\cup[7/8,7/8+1/8^{2}]\cup[1-1/8^{2},1] $
and so on... then put $K=\bigcap_{i=0}^{\infty}K_i$
[Claim: I can prove $0<H_{1/3}(K)<+\infty$ where H is the Hausdorff Measure.]
This is done by defining a sequence of functions $f_{n}$ $:[0,1]\to [0,1] $ as
$f_{0}(x)=x$
Then for n=1,2,...
$f_{n}(x)=\frac{1}{2}f_{n-1}(8x)$ for $x\in [0,1/8]$
$f_{n}(x)=\frac{1}{2} $ for $x\in [1/8,7/8]$
$f_{n}(x)=\frac{1}{2}-\frac{1}{2}f_{n-1}(8x-7)$ for $x\in [7/8,1]$
Proving these are Cauchy looking at the limit and using some theorems and results to come to the claim.
My question is then how would I prove $0<H_{2/3}(K\times K)<+\infty$
Upper bound on $H^{2/3}$
The set $K\times K$ is naturally covered by
In each of these covers, the sum of $(\operatorname{diam} Q_j)^{2/3}$, where $Q_j$ are the squares, is the same, specifically $2^{1/3}$. And for any given $\delta>0$, there is $k$ such that $8^{-k}\sqrt{2}<\delta$. So, by the definition of $H^{2/3}$, we have $H^{2/3}(K\times K) \le 2^{1/3}$.
Lower bound on $H^{2/3}$
Let $\mu$ be the restriction of the Hausdorff measure $H^{1/3}$ to the set $K$. You know that $\mu$ is positive and finite. Let $I$ be any interval of $k$th generation used in the construction of $K$; that is $|I| = 8^{-k}$. Since $K\cap I$ is just a scaled copy of $K$, scaled down by $8^{-k}$, it follows that $\mu(I) = 2^{-k} \mu(K)$. From this one can conclude that there is a constant $C$ such that $$ \mu(E) \le C(\operatorname{diam} E)^{1/3} \quad \forall E\subset K \tag1 $$ (Just cover $E$ by two suitable intervals like $I$ above.)
Consider the product measure $\nu = \mu\times \mu$, which is supported on $K\times K$. From (1) it follows that $$ \nu(E) \le C(\operatorname{diam} E)^{2/3} \quad \forall E\subset K\times K \tag2 $$ Let $(E_j)$ be any cover of $K\times K$. Then $$ \sum_j (\operatorname{diam} E_j)^{2/3} \ge C^{-1} \sum_j \nu(E_j) \ge C^{-1}\nu(K\times K) $$ Since this is true for any cover, $H^{2/3}(K\times K)\ge C^{-1}\nu(K\times K) > 0$.
Remark: both parts are quite typical for this kind of argument: an explicit cover to estimate Hausdorff measure from above; a positive measure with a growth bound (2) to estimate it from below (mass distribution principle). By combining the mass distribution principle with a deeper result, Frostman's lemma, one concludes that $$ H^t(S) > 0 \text{ and } H^s(B) > 0 \implies H^{t+s}(A\times B) $$ in general.