$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$
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).
The following is a well-known fact.
Proposition: Let $(\Omega,\mathcal{F},P)$ be a probability space.
Let $\mathcal{C}=\{X_{i}\mid i\in\Lambda\}$ be a family of random
variables (that family may be uncountable). If there exists $p\in(1,\infty)$
such that
$$
\sup_{i\in\Lambda}\int|X_{i}|^{p}\,dP<\infty,
$$
then $\mathcal{C}$ is uniformly integrable.
Proof: Let $l=\sup_{i\in\Lambda}\int|X_{i}|^{p}\,dP$. Let $q\in(1,\infty)$
be such that $\frac{1}{p}+\frac{1}{q}=1$. Let $\varepsilon>0$ be
arbitrary. Choose $c>0$ such that $l^{\frac{1}{p}}\cdot\left(\frac{l}{c^{p}}\right)^{\frac{1}{q}}<\varepsilon$,
which is possible because $l^{\frac{1}{p}}\cdot\left(\frac{l}{t^{p}}\right)^{\frac{1}{q}}\rightarrow0$
as $t\rightarrow\infty$. Define $A_{i}=\{\omega\in\Omega\mid|X_{i}(\omega)|\geq c\}$.
Observe that
\begin{eqnarray*}
l & \geq & \int_{A_{i}}|X_{i}|^{p}\,dP\\
& \geq & P(A_{i})c^{p}.
\end{eqnarray*}
That is, $P(A_{i})\leq\frac{l}{c^{p}}.$ On the other hand, by Holder
inequality,
\begin{eqnarray*}
\int_{A_{i}}|X_{i}|\,dP & = & \int|X_{i}|\cdot 1_{A_{i}}\,dP\\
& \leq & ||X_{i}||_{p}||1_{A_{i}}||_{q}\\
& \leq & l^{\frac{1}{p}}\left\{ P(A_{i})\right\} ^{\frac{1}{q}}\\
& \leq & l^{\frac{1}{p}}\left(\frac{l}{c^{p}}\right)^{\frac{1}{q}}\\
& < & \varepsilon.
\end{eqnarray*}
Therefore,
$$
\sup_{i\in\Lambda}\int_{A_{i}}|X_{i}|\,dP\leq\varepsilon
$$
and hence $\mathcal{C}$ is uniformly integrable.
Best Answer
Let $M^2=\sup_n E[|X_n|^2]$. Then $E[|X_n|\mathbb{1}(|X_n|>a)\leq \sqrt{P(|X_n|\geq a)}M\leq\frac{1}{a}M^2 $