We have a sequence $\{X_n\}_{n\ge 1}$ given (i) $EX^2_n <\infty$ for all $n\ge 1$,also (ii)$E(X_m X_n)=0$ for all $ m\not =n$
and (iii) $\sum ^ \infty _ {n=1} EX^2_n <\infty$
I was trying to show that if $S_n=\sum ^ n _ {j=1} X_j$ for all $n\ge 1$ $$$$ Then $S_n$ converges in $L^2$ to some $S$ .
$$$$
( Even though I think that I probably have to show first that $S_n=\sum ^ n _ {j=1} X_j$ is Cauchy in $L^2$ and after use the fact that $L^2$ is complete. I feel like I am missing something ).
So I started like below. $sup_{m>n} E [ (S_m-S_n)^2] = sup_{m>n} E [ (\sum ^ m _ {j=n+1} X_j)^2] = sup_{m>n} E [ (\sum ^ m _ {j,k=n+1} X_j X_k)] = sup_{m>n}\sum ^ m _ {j,k=n+1} E[ X_j X_k ] = sup_{m>n}\sum ^ m _ {j=n+1} E[ X_j^2 ]= \sum ^ \infty _ {j=n+1} E[ X_j^2 ] \to0 $ as $n\to \infty$
Hence, $S_n $is Cauchy in $L^2$ .
If I have done everything correctly how do I finish the proof? is it enough to infer that $S_n$ converges in $L^2$ to some $S$ just from the completeness of $L^2$
Your proof is correct. The point is, that the condition $E[X_iX_j] = 0$ for all $i \neq j$ ensures the identity $E[\sum_{i \in I} (X_i^2)] = \sum_{i \in I} E[X_i^2]$ for any subset $I \subset \mathbb N$ because the cross terms vanish. Once this happens, then $(S_m -S_n)^2$ is dominated in expectation by $\sum_{n=1}^\infty E[X_i^2]$ which converges to zero since the entire sum is convergent.
Note that $X_iX_j$ being independent mean zero random variables with summable variance implies the starting conditions (which are weaker : the condition is called "uncorrelated"). Furthermore, it is also not difficult to show that $S_n$ in fact converge to the same random variable $S$ almost surely (follows from bounded variance of $S_n$ implying uniform integrability).