In my course we defined the Hausdorff meaure as follows: $$ \begin{align*} \mathcal{H}_\delta^d(A)=\inf \left\{\sum_{k = 1}^{\infty } r_k^d\ \middle|\ A \subseteq \bigcup_{k \in I} B\!\left(x_k, r_k\right), 0<r_k<\delta\right\}, \quad \mathcal{H}_{ }^{ d}\!\left( A\right) = \lim_{\delta \to 0} \mathcal{H}_{ \delta }^{ d}\!\left( A\right) . \end{align*} $$ Now I want to show that $\mathcal{H}_{ }^{ n}\!\left( [ 0, 1] ^{n}\right) < + \infty $ and came up with the following: Fix any $\delta > 0$. We can cover $[ 0, 1] ^{n}$ by means of rectangles with side lengths $\sqrt{ 2} \delta $. Writing $m:= \left\lceil ( \sqrt{ 2} \delta ) ^{-1}\right\rceil $ we obtain $$ \begin{align*} [ 0, 1] ^{n} \subset \bigcup_{k _{ 1} = 0}^{m -1} \cdots \bigcup_{k _{ m} = 0}^{m - 1} [k _{ 1}\sqrt{2}\delta , ( k _{ 1} + 1) \sqrt{2}\delta ] \times \cdots \times [ k _{ m}\sqrt{2}\delta , ( k _{ m } + 1) \sqrt{2}\delta ] .\end{align*} $$ Now, each of these rectangles can in turn be covered by balls of radius $\delta $ (for every rectangle we take one of these balls). This gives us $$ \begin{align*} \mathcal{H}_{ \delta }^{ d}\!\left( A\right) \le m ^{n} \delta ^{n} \le 2^{n / 2}< +\infty \implies \mathcal{H}_{ }^{ d}\!\left( A\right) < +\infty .\end{align*} $$
My question: Is there a simpler way to show this?
Your argument just boils down to subdividing $[0,1]^n$ into sufficiently small boxes and covering each with a small ball, which in my opinion is probably the most straightforward and intuitive way to approach the problem, and your argument is still very short and simple.
The only "simpler" proof I can think of is to simply use the fact that $\mathcal{H}^n=\omega_n\mathcal{L}_n$ for some $\omega_n>0$, where $\mathcal{L}_n$ is the Lebesgue measure on $\mathbb{R}^n$. Then it is as simple as noting that
$$\mathcal{H}^n([0,1]^n)=\omega_n\mathcal{L}_n([0,1]^n)=\omega_n<\infty.$$
This does, however, require a bit more work in that you need to prove the above equality first, and so you don't get a short and simple proof from just the definition using this approach.