GIVEN $P(A)>0$ and $P(B)>0$.
If $A$ and $B$ are independent, then they cannot be mutually exclusive.
My proof:
Let $A$ and $B$ be two independent sets such that $\mathbb{P}(A)>0$ and $\mathbb{P}(B)>0 $. $$\Rightarrow \mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)>0$$ Now suppose that $A$ and $B$ are mutually exclusive, then $$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=0$$ But this is a contradiction, as $\mathbb{P}(A)\mathbb{P}(B)>0$ by design. Hence $A$ and $B$ cannot be mutually exclusive.
If $A$ and $B$ are mutually exclusive, then they cannot be independent.
My proof:
Let $A$ and $B$ be two sets such that $\mathbb{P}(A)>0$ and $\mathbb{P}(B)>0 $. If $A$ and $B$ are mutually exclusive, then $$\mathbb{P}(A\cap B)=0$$ Now suppose $A$ and $B$ are independent, then $$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=0$$ $$\Rightarrow \mathbb{P}(A)=0 \ \ \ \text{and/or} \ \ \ \mathbb{P}(B)=0$$ But this is a contradiction, as $\mathbb{P}(A)>0$ and $\mathbb{P}(B)>0$ by design. Hence $A$ and $B$ cannot be independent.
Are these proofs correct? I have tried to make them simple, but are they too simple that they fail to prove each relevant statement?
The proofs are correct, but as a matter of style, I would recommend saying "Now suppose [...] by contradiction" to emphasize that you're doing a proof by contradiction.