$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$
using Jensen.
You didn't apply Jensen's inequality correctly; it should read
$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} \left( E[|X_n|^p] \right)^{\color{red}{\frac{1}{p}}}.$$
[...] and the claim follows by letting $M \rightarrow \infty$.
No, it's not that simple. Letting $M \to \infty$ you get
$$\lim_{M \to \infty} \sup_n \mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \sup_{n \in \mathbb{N}} \|X_n\|_p,$$
but that's not good enough; you have to show that the limit equals $0$. Hint for this problem: Use Markov's inequality, i.e.
$$\mathbb{E}(|X_n| 1_{\{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p 1_{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p).$$
Define $$M_0:=\max_{n \in N} |X_n|.$$ Then we have $$E[|X_n| 1_{|X_n|>M_0}]= E[|X_n|\cdot 0 ] = 0,$$
No this doesn't work, because $M_0$ depends on $\omega$. Unfortunately, this means that your approach fails. Hint for this one: Using e.g. the dominated convergence theorem check first that the set $\{f\}$ is uniformly integrable. Extend the approach to finitely many integrable random variables.
When $E[\sup_n |X_n|] < \infty$, then the sequence is uniformly integrable.
Hint: By assumption, $Y := \sup_n |X_n|$ is integrable and $|X_n| \leq Y$ for all $n \in \mathbb{N}$. Consequently,
$$\mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \mathbb{E}(|Y| 1_{|Y|>M}) \qquad \text{for all $M>0$ and $n \in \mathbb{N}$.}$$
Now use the fact that $\{Y\}$ is uniformly integrable (see question nr. 2).
Assume for simplicity that $\mathsf{P}(A_i)>0$ for all $i$. Every element of $\mathcal{A}$ is an at most countable union of sets in $(A_n)$. Thus, for $B=\bigcup_{i}A_i$,
$$
\mathsf{E}X1_B=\sum_{i}\mathsf{E}[X1_{A_i}]=\sum_{i}\mathsf{E}[\mathsf{E}[X\mid A_i]1_{A_i}]=\mathsf{E}\left[\sum_i\mathsf{E}[X\mid A_i]1_{A_i}1_B\right]
$$
because $(A_n)$ are disjoint and $X$ is bounded (although, it suffices to assume that $\mathsf{E}|X|<\infty$).
Best Answer
Oh, it is true.
Let $Y=\sum_{n=1}^\infty X_n1_{A_n}$ which is well defined because $\{A_n\}$ is a partition. It is easy to see $$Y^+=\sum_{n=1}^\infty X_n^+1_{A_n}\quad\mathrm{and}\quad Y^-=\sum_{n=1}^\infty X^{-}_n1_{A_n}.$$ Hence by monotone convergence theorem, $$EY^+=\sum_{n=1}^\infty E\big(X_n^+1_{A_n}\big)\quad\mathrm{and}\quad EY^-=\sum_{n=1}^\infty E\big(X_n^-1_{A_n}\big).$$ Because $EX<\infty$, or equivalently $EX^+<\infty$, we have $EY^+\leq EX^+<\infty$ since $X^+_n\leq X^+$ for all $n$. Therefore $$EY=EY^+-EY^-=\sum_{n=1}^\infty E\big(X_n^+1_{A_n}\big)-\sum_{n=1}^\infty E\big(X_n^-1_{A_n}\big)=\sum_{n=1}^\infty E\big(X1_{A_n}\big).$$