Proof of: If $P(A) = P(B) = 1$ then $P(A \cap B) = 1$. Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(A\cap B)= P(A) \cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1\cdot 1=1$ so $P(A\cap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks

Well, you cannot say $P(A\cap B)= P(A) \cdot P(B)$ because you do not know if $A$ and $B$ are independent. 
Anyway, you could say $$P(A\cap B) = 1-P((A\cap B)^c)=1-P(A^c \cup B^c) \geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
