if $A \subseteq B$, then $A \cap C \subseteq B\cap C$ So I tried doing this problem myself, and the answer that I got seems right, yet at the same time I feel like the way I did it is kind of.... wonky? It seems weird basically, and I was hoping someone can help me validate my answer 
proof: 
Suppose $x \in A \subseteq B$.
Then  $x \in A$  and  $x \in B$.
Thus, if $x \in A \cap C$,
Then $x\in A$  and  $x\in C$ .
Since $x \in A$ and $x \in B$ and $x \in C$,
Then $x \in B$ and $x \in C$,
which implies that $x \in B \cap C$ .
Thus $A\cap C \subseteq B \cap C$ .
Therefore, if $A \subseteq B$, then $A\cap C\subseteq B\cap C$
I'm just not sure if it was alright to assume that  $x \in A \cap C$, which is what makes me feel like my proof may be wrong and weird. 

We have that $A,B$ and $C$ are sets.

For all sets $A, B$, and $C$, if $A\subseteq B$ then $ A\cap C\subseteq B\cap C$.

From the start of your proof

Suppose $x \in A \subseteq B$.
  Then  $x \in A$  and  $x \in B$.
  Thus, if $x \in A \cap C$,   

It is not necessary to assume that $x \in A \subseteq B$. Instead, you should assume that $x\in  A\cap C$ and deduce that $x\in B\cap C$. This method is the direct proof technique.
The proof would then be
Let $A,B,$ and $C$ be sets and assume that $A\subseteq B$. We want to show that $A\cap C\subseteq B\cap C$. Let $x\in A\cap C$. Then, by the definition of intersection, we have $x\in A$ and $x\in C$. Since $x\in A$ and $A\subseteq B$, it follows from the definition of subset that $x\in B$. Therefore, we have shown that  $x\in B$ and $x\in C$. Again, by the definition of intersection, we can conclude that $x\in B\cap C$. Because $x$ was arbitrarily chosen, we can now conclude that $A\cap C\subseteq B\cap C$.  
