= Solution
Because $X$ is a martingale,
$$
\mathbb E[\xi_n\mid\mathcal F_{n-1}]=0.
$$
Writing
$$
p_n=\mathbb P(\xi_n=1\mid\mathcal F_{n-1}),
$$
and using $\xi_n\in\{-1,1\}$ gives
$$
0=p_n-(1-p_n)=2p_n-1.
$$
Thus
$$
\mathbb P(\xi_n=1\mid\mathcal F_{n-1})
=\mathbb P(\xi_n=-1\mid\mathcal F_{n-1})
=\frac12.
$$
For signs $e_1,\ldots,e_m\in\{-1,1\}$, repeated use of the <tower property of conditional expectation> gives
$$
\begin{aligned}
\mathbb P(\xi_1=e_1,\ldots,\xi_m=e_m)
&=\mathbb E\left[
\mathbf1_{\{\xi_1=e_1,\ldots,\xi_{m-1}=e_{m-1}\}}
\mathbb P(\xi_m=e_m\mid\mathcal F_{m-1})
\right]\\
&=\frac12\mathbb P(\xi_1=e_1,\ldots,\xi_{m-1}=e_{m-1})\\
&=2^{-m}.
\end{aligned}
$$
\b[Hence the increments are <independent and identically distributed random variables>[IID] symmetric signs, proving the <symmetric signs forced by the martingale property> result. Thus $X_n-X_0$ is a <simple symmetric random walk>.]
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