Let $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $. Then $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$. Let $A_1,A_2,\dots, A_n$ be sets such that $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $ holds for all $n$. Then $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.
Is the following proof correct?
Proof:
If $A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing $, then  $A_1,A_2,\dots, A_n$ must all have at least one common element. Therefore the sets  $A_1,A_2,\dots, A_n$ are all non-empty. Hence there exists one non-empty set among  $A_1,A_2,\dots, A_n$. Hence $A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$.

Writing formal details:
$$A_1 \cap A_2 \cap \dots \cap A_n \neq \varnothing \Rightarrow\\\Rightarrow \exists x \in A_1 \cap A_2 \cap \dots \cap A_n \Rightarrow\\\Rightarrow \exists i, 1 \leqslant i \leqslant n, x \in A_i \Rightarrow\\ \Rightarrow\ x \in A_1 \cup A_2 \cup \dots \cup A_n \Rightarrow\\ \Rightarrow A_1 \cup A_2 \cup \dots \cup A_n \neq \varnothing$$
