A help to understand the generalized version of the associative law of union

Your proof is fine.

Consider $I_0=\{0\}$, $I_1=\{1,2\}$, and $J=\{0,1\}$. Then $K=\{0,1,2\}$, so the result essentially says that

$$\bigcup_{k\in K}A_k=A_0\cup(A_1\cup A_2)\;.$$

If instead you set $I_0=\{0,1\}$ and $I_1=\{2\}$, it essentially says that

$$\bigcup_{k\in K}A_k=(A_0\cup A_1)\cup A_2\;.$$

Indirectly, therefore, it says that

$$A_0\cup(A_1\cup A_2)=(A_0\cup A_1)\cup A_2\;.$$

Your proof is correct. I guess one way to see why this is a generalization is to note that usually the associative law of union is stated via three sets:$$A \cup (B \cup C) = (A\cup B)\cup C.$$ We can also write it in this form: $$A_1 \cup (A_2 \cup A_3) = (A_1\cup A_2)\cup A_3.$$ The LHS corresponds to the case where $I_1 = \{1\}$, $I_2 = \{2,3\}$, and $J=\{1,2\}$, and the RHS corresponds to the case where $I_1 = \{1,2\}$, $I_2 = \{3\}$ and $J=\{1,2\}$. They are both equal to $\bigcup_{k\in K} A_k$ with $K = \{1,2,3\}$

$\cup_k A_k$ is defined in your equations to be a union of unions, with potentially infinitely many sets being unioned together in each union and potentially infinitely many unions being joined together into the final union. So the fact that the "union of unions" is equal to the union of all the constituents is a generalization of the basic associative law fact that $(A \cup B) \cup C$ = $A \cup (B \cup C)$, because you could define the union-of-unions any way you want, distributing the sets among the individual unions being unioned together, just as long as the final collection of constituent sets is the same. Hope that helps. And yes, your proof looks correct. :)