Hint: $\|X-Y\|_2\leqslant\|X_n-X\|_2+\|X_n-Y\|_2$. And some random variables $X$ and $Y$ such that $\|X-Y\|_2=0$ are such that...
Application: If $X_n\to X$ in $L^2$ and $X_n\to Y$ in $L^2$, then $\|X_n-X\|_2+\|X_n-Y\|_2\to0$ hence $\|X-Y\|_2=0$. For every positive $u$, $ (X-Y)^2\geqslant u^2\cdot[|X-Y|\geqslant u], $ hence $ \|X-Y\|_2^2\geqslant u^2\cdot\mathrm P(|X-Y|\geqslant u). $ If $\|X-Y\|_2=0$, this shows that $\mathrm P(|X-Y|\geqslant u)=0$ for every positive $u$. That is, $\mathrm P(|X-Y|\gt0)=0$, which is equivalent to $X=Y$ almost surely.
One can adapt this proof to the convergence in probability. For every positive $u$, $ [|X-Y|\geqslant2u]\subseteq[|X_n-X|\geqslant u]\cup[|X_n-Y|\geqslant u], $ hence $ \mathrm P(|X-Y|\geqslant2u)\leqslant\mathrm P(|X_n-X|\geqslant u)+\mathrm P(|X_n-Y|\geqslant u). $ Maybe you can continue...
Edit Recall that the limits almost sure, in $L^p$ and in probability are only defined almost surely, that is, if $X_n\to X$ in either one of these three acceptions and if $X=Y$ almost surely, then $X_n\to Y$ as well. If $X_n\to X$ in $L^2$ for example, use $\|X_n-Y\|_2\leqslant\|X_n-X\|_2+\|X-Y\|_2$ and the fact that $\|X-Y\|_2=0$ if (and only if) $X=Y$ almost surely. Thus $\|X_n-Y\|_2\leqslant\|X_n-X\|_2$ and $\|X_n-X\|_2\to0$, which implies that $\|X_n-Y\|_2\to0$.