= Solution
Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y=\sum_{n\geq0}|\Delta X_{n+1}|\mathbf1_{\{T>n\}}$. Since $T$ is a <stopping time>, $\{T>n\}\in\mathcal F_n$. The <Tonelli theorem>, <conditional expectation>, and (c)'s <tail-sum formula for expectation> yield
$$
\begin{aligned}
\mathbb EY
&=\sum_{n\geq0}\mathbb E\left[\mathbf1_{\{T>n\}}\mathbb E(|\Delta X_{n+1}|\mid\mathcal F_n)\right]\\
&\leq C\sum_{n\geq0}\mathbb P(T>n)=C\mathbb ET<\infty.
\end{aligned}
$$
Thus $Y$ is an <integrable random variable>. Since $X_0=0$, telescoping gives $|X_T|\leq Y$ and $|X_n|\mathbf1_{\{T>n\}}\leq Y\mathbf1_{\{T>n\}}$. The <dominated convergence theorem> gives $\mathbb E[|X_n|\mathbf1_{\{T>n\}}]\to0$, so (b) gives <uniform integrability> of the <stopped martingale>.
The <optional stopping theorem> at the bounded <stopping time> $n\wedge T$ gives $\mathbb E X_{n\wedge T}=\mathbb E X_0=0$. Passing to the limit using the <convergence in L1> proved in (b),
$$
\boxed{\mathbb E X_T=\mathbb E X_0=0.}
$$
One can also pass directly by the <dominated convergence theorem>, since $|X_{n\wedge T}|\leq Y$.
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