X, $X_1$, $X_2$, ... uniformly bounded random variables, then convergence inprobability implies convergence in quadratic mean Let X, $X_1$, $X_2$, ... uniformly bounded random variables.
Proof:
If $X_n$ converges in probability to X, then $X_n$ converges in 2-mean (quadratic mean, square mean) to X
Convergence in probability: 
If for every $\epsilon>0$, $\lim\limits_{n \to\infty} P(|X_n-X|>\epsilon)=0$, then  $X_n$ converges in probability to X.
Convergence in 2-mean: 
If $\lim\limits_{n \to\infty} E((X_n-X)^2)=0$, then  $X_n$ converges in 2-mean to X.
 A: Showing that $E(X_n-X)^{2} \to 0$ is equivalent to showing that every subsequence of $(E(X_n-X)^{2})$ has a further subsequence which tends to $0$. Since this subsequence converges in probability to $0$ it has a subsequence which converges almost surely. Now just apply Bounded Convergence Theorem. 
Here is a completely elementary proof: suppose $|X_n| \leq C$ and $|X| \leq C$. Then  $$E(X_n-X)^{2}=E(X_n-X)^{2}I_{|X_n-X|>\sqrt {\epsilon}}+E(X_n-X)^{2}I_{|X_n-X| \leq\sqrt {\epsilon}}$$ $$\leq 4C^{2} P(|X_n-X|>\sqrt {\epsilon})+\epsilon <2\epsilon$$ if $n$ is sufficiently large. 
A: Since $X, X_1, X_2, \ldots$ are uniformly bounded, $X^k, X_1^k, X_2^k, \ldots$ are uniformly integrable for $k = 1, 2, \ldots$. Hence, $X_n^k \to X^k$ in $L^1$ for $k = 1,2, \ldots$ (Recall convergence in probability and uniform integrability together are equivalent to $L^1$ convergence.)
If you then expand $(X_n - X)^2 = X_n^2 + X^2 - 2X_nX$ and apply the argument above to $X_n^2$ and $X_nX$ you finish the proof.
