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. 

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 $A\cap C\subseteq B\cap C$.
