Given,
A ⊆ B' ∪ C
⟹A ⋂ B ⊆ (B' ⋃ C) ⋂ B (Taking intersection with B on both sides of the equation)
⟹A ⋂ B ⊆ (B' ⋂ B) ⋃ (C ⋂ B) (Distributive property)
⟹A ⋂ B ⊆ ∅ ⋃ (C ⋂ B) (B' ⋂ B = ∅ )
⟹A ⋂ B ⊆ C ⋂ B (∅ ⋃ B = B)
⟹A ⋂ B ⊆ C ( as C ⋂ B ⊆ C )
Hence, proved.
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