Let $\{X_n \}$ be a sequence of random variables on a probability space $(\Omega,\mathcal{F},P)$.
Then, $\{X_n\}$ is uniformly integrable if $$\lim_{M \to \infty} \sup_n \int_{|X_n| > M } |X_n| = 0 \tag{1}$$
From this, I know that $$\displaystyle \sup_n E(|X_n|) < \infty \tag{2}$$
But, then I am wondering whether (2) implies (1). If not, can you give me a counterexample?
No, the converse is, in general, not true. Just consider $((0,1),\mathcal{B}(0,1))$ endowed with Lebesgue measure and the sequence of random variables $$X_n(x) := 2n \cdot 1_{(0,1/n)}(x).$$ Then $$\mathbb{E}(|X_n|) = 2$$ for all $n \in \mathbb{N}$, but $$\limsup_{M \to \infty} \sup_{n \in \mathbb{N}} \int_{|X_n|>M} |X_n| \, d\mathbb{P} \geq \int_{|X_M|>M} |X_M| \, d\mathbb{P} = 2.$$
Remark: If the sequence $(X_n)_{n \in \mathbb{N}}$ is bounded in $L^p$ for some $p>1$, i.e. $$\sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|^p)<\infty,$$ then $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable.