prove that $S_{\tau \land n} \to S_{\tau}$ in $L^1$ for a random walk with $E\tau^{1/2} < \infty$ 
Let $X_i$ be a sequence of iid $L^2$ RVs with $EX_i = 0$ and define the martingale $S_n = \sum_1^n X_i$. I want to show that if $\tau$ is a stopping time and $E\tau^{1/2} < \infty$, then $ES_{\tau} =0$. 

I have been given the hint that if $Y_n$ is an $L^2$ martingale, then $E(\sup_n | Y_n|) \leq 3E(\sqrt{A_{\infty}})$, where $A = \langle Y \rangle$ is the quadratic variation process of $Y$, i.e. $A_0 =0$, $$A_n = \sum_{k=1}^nE(Y_k^2 \mid \mathcal{F}_{k-1}) - E(Y_{k} \mid \mathcal{F}_{k-1})^2$$ and $\lim_n A_n =A_{\infty}$.
My work:
Following the idea in Doob's OST, clearly have that $S_{\tau \land n}$ is a martingale and therefore $E[S_{\tau \land n}] = E[S_0] = 0$. Since $\tau< \infty$ a.s, I need to show $\lim_n E[S_{\tau \land n}] = E[S_{\tau}]$ (that is, $S_{\tau \land n} \to S_{\tau}$ in $L^1$). 
I have tried to follow this hint, and noted that since $X_i$ are $L^2$, $S_n = \sum_{i=1}^n X_i \in L^2$. Also, 
$$\begin{aligned}
A_n &= \langle S \rangle_n =\sum_{k=1}^nE(S_k^2 \mid \mathcal{F}_{k-1}) -0\\
& = E \left(\sum_{k=1}^nX_k^2 + 2\sum_{i<j \leq n} X_iX_j \mid \mathcal{F}_{k-1} \right)\\
& = E(X_n^2) + \sum_{k=1}^{n-1}X_k^2 + 2 \sum_{i<j<n}X_iX_j \\
&= E(X_n^2)+S_{n-1}^2
\end{aligned}$$
I don't know that this has a limit. If I use the hint on the $X_i$s, I get that $$\langle X\rangle_n = E(X_n^2) \implies E(\sup_n|X_n|) \leq 3(X_{\infty}^2)$$ which I don't think is helpful.  
I'm not sure exactly which route I should be going down to show convergence, there are so many theorems on it with slightly different conditions.
I'm so lost, any ideas would be much appreciated, thanks!

First of all: Your calculation of $A_n$ is not correct; note that
$$A_n = \sum_{k=1}^n \mathbb{E}(S_k^2 \color{red}{-S_{k-1}^2} \mid \mathcal{F}_{k-1})$$
and then you will end up with $$A_n = \sum_{k=1}^n \mathbb{E}(X_k^2) = n \mathbb{E}(X_1^2). \tag{1}$$

If we consider the stopped process $Y_n := S_{n \wedge T}$, then it follows from $(1)$ that its compensator is given by $A_n := \mathbb{E}(X_1^2) \min\{n,T\}$. Applying the inequality which you were given in the hint, we thus find
$$\mathbb{E} \left( \sup_{n \geq 1} |S_{n \wedge T}| \right) \leq 3 \mathbb{E}(\sqrt{T}) < \infty.$$
Now it follows easily from the dominated convergence theorem that $S_{T \wedge n} \to S_T$ in $L^1$.
