Could you please check and comment my following attempt of a proof:
Let $(\Omega,\mathcal F,\Bbb P)$ be a probality space and $(E,\mathcal E)$ a measurable space. $E$ countable and $\mathcal E = 2^E$.
Proof the following:
A discrete random variable $X:(\Omega,\mathcal F,\Bbb P) \to (E,\mathcal E)$ is independent from itself $\Leftrightarrow$
$\exists x \in E,$ such that $\Bbb P[X=x]=1$ and $\Bbb P[X=y]=0 \forall y\neq x$
My attempt:
$"\Rightarrow"\\$
Assumed, there exists $x,y \in E$ such that $x \neq y$ and $\Bbb P(A_x)>0$ and $\Bbb P(A_y)>0$ with $A_x,A_y \in \mathcal E$ containing $x$ and $y$, then we have $$0=\Bbb P[\emptyset] = \Bbb P(A_x \cap A_y)= \Bbb P(A_x) \Bbb P(A_y) >0 ↯$$ $$1=\Bbb P[E] = \Bbb P( \overset{.}{\bigcup}_{x\in E}\{X=x\})= \sum_{x \in E} \Bbb P(X=x)$$
$\Rightarrow \exists !x :$ $\Bbb P[X=x]=1$ and $\forall y= x :\Bbb P[Y=y]=0 $
Because $X$ is independent from itself, it is $$\Bbb P(X \cap X)= \Bbb P(X) = \Bbb P(X) \cdot \Bbb P(X)$$ $\Rightarrow \Bbb P(X)=\Bbb P^2(X)$ and as $\Bbb P \in [0,1]$ therefore either $\Bbb P(X)=0$ or $\Bbb P(X)=1$
$"\Leftarrow"$
Let $x,y \in E$ such that $x \neq y$ with $\Bbb P(X=x)=1$ and $\Bbb P(X=y)=0$
Then we have, if $$\Bbb P(A_x)=0 \Rightarrow \Bbb P(A_x \cap A_y)=0$$ as $\Bbb P \in [0,1]$ and $A_x \cap A_y \subset A_x$ and therefore $\Bbb P(A_x \cap A_y) \leq \Bbb P(A_x)=0$
And if $$\Bbb P(A_x)=1 \Rightarrow \Bbb P(A_x \cap A_y)=\Bbb P(A_y)$$ as $A_y= (A_y \cap A_x) \overset{.}{\cup} (A_y \cap A_x^c) $ and $(A_y \cap A_x^c)=0$ because $\Bbb P(A_x^c)=0$
Therefore $\Bbb P(A_y)=\Bbb P(A_y)\cdot\Bbb P(A_x)$ analogously for $A_y$
$\Rightarrow \Bbb P(A_x \cap A_y)=\Bbb P(A_x) \Bbb P(A_y) $
Remark: My mothertongue isn't english and I'm not good at LaTex. I apologize in advance.