Prove that for all sets $A$ and $B$ $A\subseteq B$ implies $A\cap B=A$. In the next proof we use the following lemmas:
For $A$ and B sets, $A \subseteq B$ implies $A \cup B = B$.
For all sets $A$, $A \cap A = A$.
For all sets $B$, $B \cap \varnothing = \varnothing $.
Assume that $A⊆B$. As a result, $A \cup B = B$. Note that foregoing statement suggests two possible cases.
Case 1: $A = B$.
Since $A = B$, as a matter of fact we are trying to prove that $B\cap B=B$ which is true. Therefore, $A \cap B = A$ trivially.
Case 2: $A = \varnothing$
$A ∩ B = \varnothing ∩ B = B ∩ \varnothing = \varnothing = A.$
Then, 
$A \cap B = A.$
∎
Is this proof right?
 A: This can be proved using 'modus tollens', which is:

$P\to Q \implies \lnot Q \to \lnot P$

So, we need to show that:
$$A\cap  B\ne A \to A \not\subseteq B$$
If $A\cap  B\ne A$, then $\exists x\in A$, such that $x\not\in B$ which means $A \not\subseteq B$, as required.
Modus tollens then states that the contrapositive, i.e. your original statement, is also true.
A: Let me add little generalization to subject, with hope to be useful. There is not only implication, but equivalence:
$$A \subseteq B \stackrel{\text{(1)}} \Leftrightarrow A \cup B = B $$$$A\subseteq B \stackrel{\text{(2)}} \Leftrightarrow A \cap B = A$$
(1).
From beginning we consider first equivalence and suppose $A \subseteq B$, so $(x \in A \Rightarrow x \in B) $ is true. With this admission we have:
$$x \in A \lor x \in B \Leftrightarrow (x \in A \lor x \in B) \land 1 \Leftrightarrow (x \in A \lor x \in B) \land (x \in A \Rightarrow x \in B) \Leftrightarrow \\ \Leftrightarrow (x \in A \lor x \in B) \land (x \notin A \lor x \in B) \Leftrightarrow  x \in B$$
Now suppose $A \cup B = B$ and let's proof $A \subseteq B$:
$$x \in A \Rightarrow x \in A \lor x \in B \Rightarrow x \in A \cup B = B$$
(2).
Again let's take $A \subseteq B$, so $(x \in A \Rightarrow x \in B) $ is true. As we know $A \cap B \subset A$ holds always, then we consider only reverse direction. In case of our admission
$$x \in A \Leftrightarrow \left( x \in A \land (x \in A \Rightarrow x \in B) \right) \Rightarrow x \in B$$.
And, at last,suppose we have $A \cap B = A$, then will be
$$x \in A = A \cap B \subset B \Rightarrow x \in B$$
