I tried to come up with a proof on my own and I would appreciate it if someone could please verify my proof, give me any suggestions for it, and help me strenghten one step of it:
Goal: $\text{ } \ $Prove that if $A_1, A_2, ...A_n \subset \mathbb{R^d}$, then $m^*(A_1 \cup A_2 \cup \cdots \cup A_n) \le m^*(A_1)+m^*(A_2) + \cdots m^*(A_n)$
Lemmas (verification not neccesary):
$1)$ Let $A, B \subset \mathbb{R}$, and suppose $\inf A$ and $\inf B$ exist. If $A+B=\{a+b|a \in A, B \in B\}$, then $\inf (A+B) = \inf A + \inf B$.
$2)$ Let $A, B \subset \mathbb{R}$. Suppose there exists a function $f:B \to A$ such that $\forall b \in B: f(b) \le b$. Then $\inf A \le \inf B$.
Proof of the goal: We proceed by induction on $n$, the number of subsets of $\mathbb{R^d}$. First, we prove it for $n=2$:
Let $A_1, A_2\subset \mathbb{R^d}$. We wish to prove that
$$\inf \{\sum_{k \in K} Z_k | Z_k \text{ is a box of } \mathbb{R^d}, A_1 \cup A_2 \subseteq \bigcup_{k \in K} Z_k \}\le \inf \{\sum_{j \in J} X_j | X_j \text{ is a box of } \mathbb{R^d}, A_1 \subseteq \bigcup_{j \in J} X_j \} + \inf \{\sum_{i \in I} Y_i | Y_i \text{ is a box of } \mathbb{R^d}, A_2 \subseteq \bigcup_{i \in I} Y_i \}$$
By Lemma 1, the RHS is equal to
$$\inf \{\sum_{j \in J} X_j +\sum_{i \in I} Y_i| X_j, Y_i \text{ are boxes of } \mathbb{R^d}, A_1 \subseteq \bigcup_{j \in J} X_j , A_2 \subseteq \bigcup_{i \in I} Y_i \}$$
Let the set on the $LHS$ be $A$ and let this new set on the $RHS$ be $B$. Define a function $f:B \to A$ as follows: take any $b \in RHS$. $b=\sum_{j \in J} X_j +\sum_{i \in I} Y_i$ for some $S = \{X_j|j \in J \}$ and $T = \{Y_i |i \in I \}$ and cosider $W=S \cup T := \{W_k|k\in K\}$. We have that $A_1 \cup A_2 \subseteq \bigcup_{k \in K} W_k$, so $\sum_{k \in K} W_k $ is a member of the $LHS$.
$\color {red}{\text {Claim I cannot justify easily:}}$
$$\sum_{k \in K} W_k \le \sum_{j \in J} X_j +\sum_{i \in I} Y_i=b$$
Define $f(b)$ as $\sum_{k \in K} W_k$. Thus $f(b) \le b$, so by Lemma 2 $\inf LHS \le \inf RHS$, so $m^* (A_1 \cup A_2) \le m^*(A_1) + m^* (A_2)$.
Now it follows to complete the induction, but I don't need help with that.
Your proof is fine. I just find it a little confusing that you use $\sum_k Z_k$ instead of $\sum_k m(Z_k)$ where $m$ is the measure of boxes in $\mathbb{R}^d$. Also, you use the set $K$ twice, once in $Z_k$ and once in $W_k$.
For the last step, if you think about it, $\sum_j m(X_j) + \sum_i m(Y_i)$ is actually the same as $\sum_k m(W_k)$ because $W = S\cup T$. So the function $f: B\to A$ can be defined as $f(b) = b$, which obviously satisfies the hypothesis in Lemma 2.
Here we only use a special case of Lemma 2 where $f$ is an identity function. The intuition is that, if $B \subset A$, then $\inf A \leq \inf B$.