Is the following "generalized version of distributive law of sets" true? Consider a collection of sets $\mathcal{T}_{i, s_i} \subseteq \mathbb{R}^n$ with $i \in \{0,\ldots,k\} =: \mathcal{I}$ and $s_i \in \mathcal{S}_i$, where $\mathcal{I}$ is a finite set and $\mathcal{S}_i$ is an uncountable set. Then, does the following hold?
\begin{gather*}
\bigcap_{i=0}^k \bigcup_{s_i \in \mathcal{S}_i} \mathcal{T}_{i,s_i} = \bigcup_{(s_0,\ldots,s_k) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_k}\bigcap_{i=0}^k \mathcal{T}_{i,s_i}\tag{1}
\end{gather*}
Thanks very much!
PS: I have provide a proof (using mathematical induction) here, but not sure if it is rigorous enough.
Base case: For $k = 0$, $\text{LHS of}\ (1) = \bigcup_{s_0 \in \mathcal{S}_0} \mathcal{T}_{0,s_0}$ and $\text{RHS of}\ (1) = \bigcup_{s_0 \in \mathcal{S}_0} \mathcal{T}_{0,s_0}$. Thus, (1) holds for $k = 0$.
Inductive step: Assume $(1)$ holds for $k = j \geq 0$. For $k = j+1$, we have
\begin{align*}
\bigcap_{i=0}^{j+1} \bigcup_{s_i\in\mathcal{S}_i} \mathcal{T}_{i,s_i} &= \left(\bigcap_{i=0}^{j} \bigcup_{s_i\in\mathcal{S}_i} \mathcal{T}_{i,s_i}\right) \cap \left(\bigcup_{s_{j+1}\in\mathcal{S}_{j+1}} \mathcal{T}_{j+1,s_{j+1}}\right)\\
&\stackrel{(a)}{=} \left(\bigcup_{(s_0,\ldots,s_{j}) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_{j}}\bigcap_{i=0}^{j} \mathcal{T}_{i,s_i}\right) \cap \left(\bigcup_{s_{j+1}\in\mathcal{S}_{j+1}} \mathcal{T}_{j+1,s_{j+1}}\right)\\
&\stackrel{(b)}{=} \bigcup_{(s_0,\ldots,s_{j}) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_{j}} \left[\left(\bigcap_{i=0}^{j} \mathcal{T}_{i,s_i}\right) \cap \left(\bigcup_{s_{j+1}\in\mathcal{S}_{j+1}} \mathcal{T}_{j+1,s_{j+1}}\right)\right]\\
&\stackrel{(c)}{=} \bigcup_{(s_0,\ldots,s_{j}) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_{j}} \left[\bigcup_{s_{j+1}\in\mathcal{S}_{j+1}}\left(\left(\bigcap_{i=0}^{j} \mathcal{T}_{i,s_i}\right) \cap \mathcal{T}_{j+1,s_{j+1}}\right)\right]\\
&= \bigcup_{(s_0,\ldots,s_{j}) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_{j}} \left[\bigcup_{s_{j+1}\in\mathcal{S}_{j+1}}\left(\bigcap_{i=0}^{j+1} \mathcal{T}_{i,s_i}\right)\right]\\
&= \bigcup_{(s_0,\ldots,s_{j+1}) \in \mathcal{S}_0\times\cdots\times\mathcal{S}_{j+1}}\bigcap_{i=0}^{j+1} \mathcal{T}_{i,s_i},
\end{align*}
where $(a)$ follows from $(1)$; $(b)$ and $(c)$ are established by the distributive law of sets that
$$(\mathcal{A}_1\cup\cdots\cup\mathcal{A}_m) \cap \mathcal{B} = (\mathcal{A}_1\cap\mathcal{B}) \cup \cdots \cup (\mathcal{A}_m\cap\mathcal{B}).$$
Thus, $(1)$ holds for $k = j+1$.
 A: Your inductive proof is correct, and here's anther intuitive way to prove by examining exactly what elements are in the expression on both sides.
For the LHS,\begin{gather*}
A := \bigcap_{i = 0}^k \bigcup_{s_i \in S_i} T_{i, s_i} = \bigcap_{i = 0}^k \{t \mid \exists s_i \in S_i\colon\ t \in T_{i, s_i}\}\\
= \{t \mid \forall 0 \leqslant i \leqslant k,\ \exists s_i \in S_i\colon\ t \in T_{i, s_i}\}.
\end{gather*}
For the RHS,\begin{gather*}
B:= \bigcup_{(s_0, \cdots, s_k) \in S_1 × \cdots × S_k} \bigcap_{i = 0}^k T_{i, s_i} = \bigcup_{(s_0, \cdots, s_k) \in S_1 × \cdots × S_k} \{t \mid \forall 0 \leqslant i \leqslant k\colon\ t \in T_{i, s_i}\}\\
= \{t \mid \exists (s_0, \cdots, s_k) \in S_1 × \cdots × S_k,\ \forall 0 \leqslant i \leqslant k\colon\ t \in T_{i, s_i}\}.
\end{gather*}
Now for a fixed $t \in A$, there exists $s_{i, t} \in S_k$ for any $0 \leqslant i \leqslant k$ such that $t \in T_{i, s_{i, t}}$, thus$$
(s_{0, t}, \cdots, s_{k, t}) \in S_0 × \cdots × S_k, \quad t \in T_{i, s_{i, t}}\ (\forall 0 \leqslant i \leqslant k),
$$
which implies that $t \in B$. Hence $A \subseteq B$. For a fixed $t \in B$, there exists $(s_{0, t}, \cdots, s_{k, t}) \in S_0 × \cdots × S_k$ such that $t \in T_{i, s_{i, t}}$ for any $0 \leqslant i \leqslant k$, thus$$
s_{i, t} \in S_i, \quad t \in T_{i, s_{i, t}}, \quad \forall 0 \leqslant i \leqslant k
$$
which implies that $t \in A$. Hence $B \subseteq A$, and thus $A = B$.
