Proof: Let \(x \in (A-B) \cup (B-C)\). Then by definition of union, \(x \in (A-B)\) or \(x \in (B-C)\). So we have the following cases:

- Case 1: \(x \in (A-B)\). Then by definition of set difference, \(x \in A\) and \(x \not\in B\). Since \(A \subseteq A \cup B\), \(x \in A\) implies that \(x \in A \cup B\). Since \(A \cap B \cap C \subseteq B\), \(x \not\in B)\) implies that \(x \notin A \cap B \cap C\).
- Case 2: \(x \in (B-C)\). Then \(x \in B\) and \(x \not\in C\). Since \(B \subseteq A \cup B\), \(x \in B\) implies that \(x \in A \cup B\). Since \(A \cap B \cap C \subseteq C\), \(x \not\in C\) implies \(x \notin A \cap B \cap C\).
So, in both cases, we have \(x \in A \cup B\) and \(x \notin A \cap B \cap C\). So \(x \in (A \cup B) - (A \cap B \cap C)\).

To summarize, we picked an arbitrary element x from \((A-B) \cup (B-C)\) and we have shown that it is also a member of \((A \cup B) - (A \cap B \cap C)\). By the definition of subset, this means that \((A-B) \cup (B-C) \subseteq (A \cup B) - (A \cap B \cap C)\).