Result: $(A\times C)-(B\times C)\subseteq (A-B)\times C$
Proof: Let $(x,y)\in (A\times C)-(B\times C)$. Then $(x,y)\in A\times C,\implies x\in A,y\in C$. Since $(x,y)\notin B\times C, x\notin B$. Thus $x\in A-B$ and hence $(x,y)\in (A-B)\times C$
Result: $(A\times C)-(B\times C)\subseteq (A-B)\times C$
Proof: Let $(x,y)\in (A\times C)-(B\times C)$. Then $(x,y)\in A\times C,\implies x\in A,y\in C$. Since $(x,y)\notin B\times C, x\notin B$. Thus $x\in A-B$ and hence $(x,y)\in (A-B)\times C$
The proof is in principle correct. However, one might add that $(x,y)\notin B\times C$ gives you immediately only $x\notin B\lor y\notin C$. But together with $y\in C$ (from $(x,y)\in A\times C$), you get indeed that $x\notin B$.
If one really wants to make that proof very explicit, one may werite:
Let $z\in (A\times C)-(B\times C)$ be an arbitrary element. Then $z\in A\times C$ and $z\notin B\times C$, hence there exist $x\in A$, $y\in C$ with $z=(x,y)$. If we assume $x\in B$, then this implies $z=(x,y)\in B\times C$, contrary to $z\notin B\times C$. Hence $x\notin B$, i.e. $x\in A-B$ and together with $y\in C$, we have $z=(x,y)\in(A-B)\times C$. Since $z$ was arbitrary, we conclude $(A\times C)-(B\times C)\subseteq (A-B)\times C$.