From various sources (e.g. Wikipedia), I read convergence in mean implies convergence in probability. However, I have some examples that seem to disagree with this statement.
Let $X^{(1)}$ and $X^{(2)}$ be two distinct ($X^{(1)} \ne X^{(2)}$) standard normal random variables. Let $U$ and $V$ be two random variables with bounded means. Then define $Z_n^{(1)}:=X^{(1)}+U/n$ and $Z_n^{(2)}:=X^{(2)}+V/n$.
Apparently $\lim_{n \rightarrow \infty}\mathbb{E}[|Z_n^{(1)}-Z_n^{(2)}|]=0$ but $Z_n^{(1)} \not\rightarrow _p Z_n^{(2)}$, since $X^{(1)} \ne X^{(2)}$.
I'm sure I'm stupidly wrong but could some one let me know where did I make the mistake exactly.
You are not careful enough with absolute values: $|E(X)|$ and $E(|X|)$ are very different things. In general the latter is larger than or equal to the former.
$E(|X|)$ being small is a very strong condition. In your case what you have is $$\lim_{n \rightarrow \infty}|\mathbb{E}[Z_n^{(1)}-Z_n^{(2)}]|=0$$ but there is no way to take the absolute values inside, unless, precisely, if $P(X^{(1)}=X^{(2)})=1$.