My problem: Let $(\xi_n)_{n \geq 1}$ be a sequence of bounded independent and identically distributed random variables and $S_n = \sum_{k=1}^{n} \xi_k$. Let $\mathcal{F}_n = \sigma (\xi_k : k \leq n)$ and $\tau$ a stopping time such that $\mathbb{E}[\tau]< +\infty$. Then if $\sigma^2= \mathbb{E}[(\xi_1 - \mathbb{E}[\xi_1])^2]$ it holds $\mathbb{E}[(S_{\tau}-\tau \mathbb{E}[\xi_1])^2]=\mathbb{E}[\tau] \sigma^2$.
My attempt: I defined $\overline{S}_n=S_n - n\mathbb{E}[\xi_1]$ and I noticed that it is a martingale. Suppose now that $\tau$ is bounded. Since the predictable quadratic variation of $\overline{S}_n$ is $<\overline{S}_n>_n=n\sigma^2$. Thus: $$\mathbb{E}[\overline{S}_{\tau}^2]=\mathbb{E}[<\overline{S}>_{\tau}]=\mathbb{E}[\tau]\sigma^2$$
Now I want to generalize to the general case of $\tau$ not bounded. I get: $$\mathbb{E}[\overline{S}_{\tau \wedge n}^2]=\mathbb{E}[\tau \wedge n]\sigma^2$$ I am not sure if I can pass to the limit in the LHS, i.e. in $\mathbb{E}[\overline{S}_{\tau \wedge n}^2]$.
Remark: a related question I found is this.
Since $\tau < \infty$ a.s., $(\overline{S}_{\tau \land n})_{n\in\mathbb{N_0}}$ is an $L^2$ martingale converging almost surely to $\overline{S}_{\tau}$. You have shown that $$\sup_{n \in \mathbb{N}}E(\overline{S}_{\tau \land n}^2) = E(\tau)\sigma^2 < \infty.$$ By Doob's $L^p$ convergence theorem for martingales (an easy consequence of Doob's $L^p$ inequality), $\overline{S}_{\tau \land n} \to \overline{S}_{\tau}$ in $L^2$. This lets you take limits to prove the result.