Many texts (Fremlin, Stroock,etc.) on measure theory seem to think this is a difficult lemma to prove. Tao proves this in a indirect way by approximating volume with the number of grid points.
If $J_1, \dots , J_n$ are half-open covering $I$, then $\mu(I) \le \sum\limits_{i=1}^{n} {\mu(J_i)}$. (where $\mu$ is the standard volume in $\mathbb{R}^n$)
But I have a proof in mind which is very clear and short, so I would like to know if there are issues with it.
First, any nonempty half-open cube $K$ is uniquely written as $[a,b) := \{x|a \le x < b\}$ where inequality is component-wise. (I call $l(K):=a$ the left endpoint and $r(K):=b$ the right)
Take $K_i = J_i \cap I$. Then $\mu(K_i) \le \mu(J_i)$ since $l(J_i) \le l(K_i) \le r(K_i) \le r(J_i)$
Now, let $L_j := \{l(K_i)_j\} \cup \{r(K_i)_j\} \cup \{l(I),r(I)\}$ where for example $l(K_i)_j$ is the $j$th component of $l(K_i)$. (Basically I'm projecting the cubes on each component)
Order each $L_j$ and consider the half-open cubes $[((L_1)_{a_1},\dots,(L_n)_{a_n}),((L_1)_{a_1+1},\dots,(L_n)_{a_n+1}))$ where each $a_j \le \#(L_j)-1$ (basically I'm considering the paritition into subcubes based on paritions)
Now $\mu(I)$ is the sum of the volume of the subcubes. But each subcube lies entirely in some $K_i$
(if $x$ and $y$ are in the same subcube and $x \in K_i$ then $\forall j$, $x_j \in [l(K_i)_j,r(K_i)_j)$, suppose $y_j \not\in [l(K_i)_j,r(K_i)_j)$ then there is an element of $L_j$ between two adjacent elements, so contradiction)
And $\mu(K_i)$ is the sum of the volume of the subcubes in it, so we are done.
This might look long but it's a simple idea, is it missing any important steps?