(i) In fact, even a sequence $(X_n)_{n\geqslant 1} $ which is bounded in $\mathbb L^1$ (that is, $\sup_n\mathbb E|X_n| <\infty$) is tight. This can be seen from the inequality
$$\mathbb P\{|X_n|\gt R \}\leqslant \frac 1R\mathbb E[|X_n|]\leqslant \frac 1R\sup_j\mathbb E[|X_j|].$$
(ii) When the underlying measure space is not finite, we rather use the following definition of uniform integrability: the family $\left(f_i\right)_{i\in I}$ is uniformly integrable if for each positive $\varepsilon$, there exists an integrable function $g$ such that $\sup_{i\in I}\int_{\{|f_i|\geqslant g }|f_i|<\varepsilon$.
With the classical definition in the context of probability space, we would find that the sequence $ (\mathbf 1_{(0,n)})_{n\geqslant 1}$ is uniformly integrable, which is not acceptable because it is not even bounded in $\mathbb L^1$.
$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$
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).
Best Answer
Your proof that $\mathbb E\left[Y_n\right]$ can be bounded independently of $n$ and that a sequence which is bounded in $\mathbb L^2$ is uniformly integrable is correct.