TITLE: Prove if $A ⊆ B$, then $A ∩ C ⊆ B ∩ C$.
QUESTION [5 upvotes]: For any sets $A, B$, and $C$
Assume $A ⊆ B$, and suppose, $x\in (A∩  C)$.
Then $x\in A $  and  $x\in C$ by  definition  of  $A ∩ C$.
Since $A ⊆ B$ it  follows  that  if  $x\in A$  then $x\in B$.
Thus, $x\in A$ and  $x\in C$ implies $x\in B$  and  $x\in C$.
Therefore, $x\in B ∩ C$.
FINE?
Please critique and give any advice, thank you!

REPLY [3 votes]: It looks technically correct, but I think polishing it up would do it some good. Below I give what I think is a fairly decent way to write up your argument in a verbal fashion (much as you have done), and I then provide a slightly different way (maybe you could call it a more "algebraic" approach). 


Prove that if $A\subseteq B$, then $A\cap C\subseteq B\cap C$.

Proof. Suppose $A\subseteq B$. If $x\in A\cap C$, then $x\in A$ and $x\in C$ by the definition of set intersection. Since $x\in A$, then $x\in B$ since $A\subseteq B$. Thus, if $x\in A\cap C$, then $x\in B\cap C$, as desired. $\blacksquare$
Proof. Suppose $A\subseteq B$. Then
\begin{align}
x\in A\cap C &\implies \text{$x\in A$ and $x\in C$}\tag{by definition}\\[1em]
&\implies x\in A\ \text{and}\ x\in B\ \text{and}\ x\in C\tag{$A\subseteq B$}\\[1em]
&\implies x\in B\ \text{and}\ x\in C\tag{simplification}\\[1em]
&\implies x\in B\cap C.\blacksquare \tag{by definition}
\end{align}