= Solution
Use $\mathcal F_m=\sigma(\xi_1,\ldots,\xi_m)$ and $S_0=0$. <Independence> and zero <means> imply $\mathbb E[S_{m+1}\mid\mathcal F_m]=S_m$, so the partial sums form a <square-integrable> <martingale>. More explicitly,
$$
\mathbb E[S_{m+1}^2\mid\mathcal F_m]=S_m^2+\mathbb E\xi_{m+1}^2\geq S_m^2.
$$
Thus $S_m^2$ is a nonnegative <submartingale>. The <Doob maximal inequality> at threshold $x^2$ now proves the <Kolmogorov maximal inequality>:
$$
\boxed{\mathbb P\left(\max_{0\leq m\leq n}|S_m|\geq x\right)
=\mathbb P\left(\max_{m\leq n}S_m^2\geq x^2\right)
\leq\frac{\mathbb ES_n^2}{x^2}.}
$$
Finally, expanding the terminal square and using <independence> and the zero <means> gives $\mathbb ES_n^2=\sum_{j=1}^n\mathbb E\xi_j^2$. Identical distributions are not required.
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