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. 
 A: You did a 'beginners mistake' and started with an assumption.
Namely that $A\subseteq B$. 
You should start with what you have to show. That is $A\cap C\subseteq B\cap C$ under the assumption that $A\subseteq B$.
Then the proof reads like this:
Let $x\in A\cap C$. We have to show that $x\in B\cap C$.
Since $x\in A\cap C$, we have $x\in A$ and $x\in C$. Since $x\in A\subseteq B$, it is $x\in B$. So $x\in B$ and $x\in C$. Thus $x\in B\cap C$.
It is ok to assume that $x\in A\cap C$. That is exactly what we have to do, when we want to show some 'subset relationship'.
A: Remember that $X\cap Y = X$ iff $X\subseteq Y$. According to this result and the additional hypothesis, one has
\begin{align*}
(A\cap C)\cap (B\cap C) = (A\cap B\cap C) = A\cap C \Longrightarrow A\cap C \subseteq B\cap C
\end{align*}
A: You need to start with $$x\in A\cap C$$ and show that $$x\in B\cap C$$
Let $$x\in A\cap C$$ then $$x\in A \text { and } x\in C$$
Since $x\in A$ and $A\subseteq B$ then $x\in B$
$$x\in B \text { and } x\in C \implies x\in B\cap C$$
A: The "$x$" you are referring to in lines 1 and 2 are a different "$x$" than you have in the rest of the proof.  And don't care about the $x\in A$ specifically but just that it leads to a general conclusion that we will use for the later $x$.
If I were to edit your proof but leave your thought process and pacing completly in tact, but clarify when we are making general from specific cases I'd do:
We are presuming $A\subseteq B$.
So for any $y \in A$ we'd have $y$ is $A$ and  $y \in B$.
Let $x$ be an arbitrary element in $A\cap C$.
Then x∈A and x∈C .
Since x∈A thus $x\in A$ and $x \in B$.
So x∈B and x∈C,
which implies that x∈B∩C .
Thus any element of $x \in A\cap C$ is in $B\cap C$.
Thus A∩C⊆B∩C .
Therefore, if A⊆B, then A∩C⊆B∩C
.....
But you don't need to be quite so stiff a repetitive.
It'd be enough to say.
For any $x \in A\cap C$ we have $x\in A$ and $x\in C$.
Since $A\subseteq B$ and $x \in A$ we know $x \in B$.
So $x \in B$ and $x \in C$.
So $x\in B\cap C$.
Thus $A\cap C\subseteq B\cap C$.
A: 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 $A\cap C\subseteq B\cap C$.  
