I do not really know how to prove the following statement:
If $E(|X-Y|)=0$ then $P(X=Y)=1$.
The main problem is how to handle the absolute value $|X-Y|$. If I say that $|X-Y| \geq 0$ it follows that $E(X)=E(Y)$ which is also the result for $|X-Y|<0$. But then the expectation values are equal and you can show that $P(X=Y)=1$ is not true for all $X,Y$.
On
$$ \begin{eqnarray*} 0{}={}\mathbb{E}\left[\ \big| X-Y\big|\ \right]&{}={}&\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]P\left(X\ne Y\right){}+{}\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X= Y\right]P\left(X= Y\right)\newline &&\newline &{}={}&\mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]P\left(X\ne Y\right)\,. \end{eqnarray*} $$
Therefore,
$$ \mathbb{E}\left[\ \big| X-Y\big|\ \bigg| \ X\ne Y\right]\ne0\implies P\left(X\ne Y\right)=0\implies P\left(X= Y\right)=1\,. $$
Simple proof by contradiction:
Assume that $P(X=Y)<1$, then there has to exists $x',y'$ such that $x'\neq y'$ and $P(X=x'\wedge Y=y')>0$.
But then if we look at the random variable $Z=|X-Y|$, we get that $$E(Z)=\sum_{x,y}P(X=x\wedge Y=y)\cdot|x-y|$$ $$=P(X=x'\wedge Y=y')\cdot|x'-y'|+\sum_{x,y\neq(x',y')}P(X=x\wedge Y=y)\cdot|x-y|$$
Since $$P(X=x'\wedge Y=y')\cdot|x'-y'|$$ is strictly positive and
$$\sum_{x,y\neq(x',y')}P(X=x\wedge Y=y)\cdot|x-y|$$ Is non-negative, it follows that $E(Z)>0$.