I suppose you're supposed to prove that set intersection is distributive across a union,
[tex]A\cap(B\cup C)=(A\cap B)\cup(A\cap C)[/tex]
Two sets are equal if they are subsets of one another. To prove a set [tex]X[/tex] is a subset of another set [tex]Y[/tex], you have to show that any element [tex]x\in X[/tex] also belongs to [tex]Y[/tex].
Let [tex]x\in A\cap(B\cup C)[/tex]. By definition of intersection, both [tex]x\in A[/tex] and [tex]x\in B\cup C[/tex]. By definition of union, either [tex]x\in B[/tex] or [tex]x\in C[/tex]. If [tex]x\in B[/tex], then clearly [tex]x\in A\cap B[/tex]; if [tex]x\in C[/tex], then [tex]x\in A\cap C[/tex]. Either way, [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Hence [tex]A\cap(B\cup C)\subseteq(A\cap B)\cup(B\cap C)[/tex].
The proof in the other direction uses the same sort of reasoning. Let [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Then either [tex]x\in A\cap B[/tex] or [tex]x\in A\cap C[/tex]. If [tex]x\in A\cap B[/tex], then both [tex]x\in A[/tex] and [tex]x\in B[/tex]; if [tex]x\in A\cap C[/tex], then both [tex]x\in A[/tex] and [tex]x\in C[/tex]. So certainly [tex]x\in A[/tex], and either [tex]x\in B[/tex] or [tex]x\in C[/tex] so that [tex]x\in B\cup C[/tex]. Hence [tex](A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C)[/tex].
Both sets are subsets of one another, so they are equal.
