Q:

Proving $|A-B| = |A| - |A\cap B|$

Let $U$ be the universal set. 
I can see this with the aid of a venn diagram.   
However, what is wrong with this? 
$$|A-B|:= |A\cap B^{\complement}|\\ = |A- (A\cap B)|\\ := |A\cap (A\cap B)^{\complement}| \\ = |A| + |(A\cap B)^{\complement}| - |A\cup(A^{\complement} \cup B)| \\ = |A| + |(A\cap B)^{\complement}| - |U| $$
which can't be true since in general,
$$|U| \neq |A| + |(A\cap B)^{\complement}|- |A-B|...$$

A:

I don't think there's anything that's really wrong with that, only that it didn't end up with what you tried to prove. 
As for your comment in the end I'd like to see a case where
$$|U| \ne |A| + |(A\cap B)^C| - |A-B|$$
The general approach is to decompose the sets into a disjoint partition of the universe (that is in terms of the sets $A\cap B$, $A\cap B^C$, $A^C\cap B$ and $A^C\cap B^C$).  You then have
$$|A| = |(A\cap B)\cup(A\cap B^C)| = |A\cap B| + |A\cap B^C|$$
$$|(A\cap B)^C| = |(A\cap B^C)\cup(A^C\cap B)\cup(A^C\cap B^C)| = |(A\cap B^C)|+|(A^C\cap B)|+|(A^C\cap B^C)|$$
$$|A-B| = |A\cap B^C|$$

