Prove that $ (A_1 \cap \dots \cap A_n) \triangle (B_1 \cap \dots \cap B_n) \subset (A_1 \triangle B_1) \cup \dots \cup (A_n \triangle B_n) $ Prove that $ (A_1 \cap \dots \cap A_n) \triangle (B_1 \cap \dots \cap B_n) \subset (A_1 \triangle B_1) \cup \dots \cup (A_n \triangle B_n) $ is true for any sets $A_1, \dots , A_n$ and $B_1, \dots , B_n  $
I tried to solve it using math induction.
n = 1: $A_1 \triangle B_1 \subset A_1 \triangle B_1$ is true
n = m: $ (A_1 \cap \dots \cap A_m) \triangle (B_1 \cap \dots \cap B_m) \subset (A_1 \triangle B_1) \cup \dots \cup (A_m \triangle B_m) $
n = m + 1: $ (A_1 \cap \dots \cap A_m) \triangle (B_1 \cap \dots \cap B_m) \cup A_{m+1} \triangle B_{m+1} \subset (A_1 \triangle B_1) \cup \dots \cup (A_{m+1} \triangle B_{m+1})$
But I have no idea what to do next

You can prove it directly by element-chasing; using induction just overcomplicates matters. Suppose that $x\in\left(\bigcap_{k=1}^nA_k\right)\triangle\left(\bigcap_{k=1}^nB_k\right)$; then either $x\in\left(\bigcap_{k=1}^nA_k\right)\setminus\left(\bigcap_{k=1}^nB_k\right)$, or $x\in\left(\bigcap_{k=1}^nB_k\right)\setminus\left(\bigcap_{k=1}^nA_k\right)$. Without loss of generality we may assume that $x\in\left(\bigcap_{k=1}^nA_k\right)\setminus\left(\bigcap_{k=1}^nB_k\right)$. Then $x\in\bigcap_{k=1}^nA_k$, so $x\in A_k$ for $k=1,\ldots,n$, and $x\notin\bigcap_{k=1}^nB_k$, so there is an $\ell\in\{1,\ldots,n\}$ such that $x\notin B_\ell$. But then $x\in A_\ell\setminus B_\ell\subseteq A_\ell\triangle B_\ell\subseteq\bigcup_{k=1}^n(A_k\triangle B_k)$, and since $x$ was an arbitrary element of $\left(\bigcap_{k=1}^nA_k\right)\triangle\left(\bigcap_{k=1}^nB_k\right)$, we conclude that $\left(\bigcap_{k=1}^nA_k\right)\triangle\left(\bigcap_{k=1}^nB_k\right)\subseteq\bigcup_{k=1}^n(A_k\triangle B_k)$, as desired.
