It is well-known that the Hausdorff $n$-measure (with suitable normalization) agrees with the Lebesgue measure on Euclidean $n$-space. Thus, it's clear that the $H^2$ measure of the unit square is $1$.
However, I cannot find or create a direct proof of this fact from the definition of Hausdorff measure as the limit of the Hausdorff content:
$$ H^{d}_{\delta} (S) := \inf \left\{ \sum_{i=1}^{\infty} \frac{\omega_d}{2^d}(\operatorname{diam} U_i )^d : \bigcup_{i=1}^{\infty} U_i \supseteq S, \operatorname{diam} U_i < \delta \right\}.$$
I have tried covering with squares and with circles so far, but so far the lowest Hausdorff content I can find is $\pi/2$, which is the Hausdorff content of a covering made of $2^n$ smaller squares.
To clarify the question: find a sequence of covers $\{U_i\}$ of $S$ with so that the diameter of the sets in each cover approaches zero and the sum $\sum_{i=1}^{\infty} \frac{\omega_d}{2^d}(\operatorname{diam} U_i )^d$ approaches 1.
Note: $\omega_d$ is the volume of the $d$-dimensional unit ball. A source proving agreement of this normalization with Lebesgue measure can be found in these lecture notes from CUHK professor Kai-Seng Chou: https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.math.cuhk.edu.hk/course_builder/1415/math5011/MATH5011_Chapter_3.2014.pdf&ved=2ahUKEwj975GE_7rxAhWMMd8KHSETDycQFnoECBIQAQ&usg=AOvVaw2ak7RZJQW-6f6Gth2OTBoc. See pages 14 and 18.
Best Answer
We work with the open unit square $(0,1)^2$ for ease.
You basically have to use circles as the $U_i$'s, since they are the only shape with $\frac{\pi}{4}\text{diam}(U)^2 = \text{Area}(U)$.
First, take $n \in \mathbb{N}$ large and even so that $\frac{2}{n} < \delta$. Then make a grid of $\frac{n^2}{4}$ circles each of diameter $\frac{2}{n}$ with centers at $(\frac{2j+1}{n},\frac{2k+1}{n})$ for $0 \le j,k \le \frac{n}{2}-1$. Let $\mathcal{U}^{(1)}$ be the collection of these circles.
Perform the following algorithm. For each $t \ge 1$, let $C_1^{(t)},\dots,C_{m_t}^{(t)}$ be the connected components of $[0,1]^2\setminus \mathcal{U}^{(t)}$. For each $1 \le j \le m_t$, choose the largest circle $D_j^{(t)}$ that can fit completely inside $C_j^{(t)}$. Let $\mathcal{U}^{(t+1)} = \mathcal{U}^{(t)}\cup\bigcup_{i=1}^{m_t} D_j^{(t)}$.
Let $\mathcal{U} = \cup_{t \ge 1} \mathcal{U}^{(t)}$. Now, $\mathcal{U}$ is countable, so we may write it as $\mathcal{U} = \{U_1,U_2,\dots\}$.
Fix $\epsilon > 0$. Let $\mathcal{U}^{(\epsilon)} = \{U_1^{(\epsilon)},U_2^{(\epsilon)},\dots\}$, where $U_j^{(\epsilon)}$ is the same circle as $U_j$ except with $(1+\epsilon)$ times the radius. Then $\cup_{j \ge 1} U_j^{(\epsilon)}$ covers $(0,1)^2$.
So, $1 \ge \sum_{j \ge 1} \text{Area}(U_j^{(\epsilon)}) = (1+\epsilon)^2\sum_{j \ge 1} \text{Area}(U_j)$. So, $\sum_{i \ge 1} \frac{\pi}{4}\text{diam}(U_i^{(\epsilon)})^2 = \sum_{i \ge 1} \frac{\pi}{4}\frac{4}{\pi}\text{Area}(U_i^{(\epsilon)}) \le \frac{1}{(1+\epsilon)^2}$.
As $\epsilon$ was arbitrary, the proof is complete.