Prove that process is uniformly integrable 
Let $(X_t)_{t\ge 0}$ be a stochastic process, and let $Y$ be an
  integrable random variable, such that $|X_t|\le Y$ for $t\ge0$. Prove
  that $(X_t)_{t\ge 0}$ is uniformly integrable.

From definition, we have that $(X_t)_{t\ge 0}$ is uniformly integrable if 
$$\sup_{t\in [0,\infty)}\int_{\{X_t > \epsilon \}} |X_t|d\mathbb{P} \rightarrow 0$$ when $\epsilon\rightarrow \infty$.
I think I can use here Dominated convergence theorem.
Define a sequence of functions: $Y_t^{\epsilon}:=X_t \mathbb{1}_{\{X_t > \epsilon \}}$
Because $|X_t|$ is bounded by $Y$ and $Y$ is integrable, we have that $(Y_t^{\epsilon})_{\epsilon} \rightarrow 0$
From Dominated convergence theorem we have that $\int_{\{X_t > \epsilon \}} |X_t|d\mathbb{P} \rightarrow 0$ as $\epsilon \rightarrow \infty$
I'm very doubtful about my solutions, itts probably wrong. Can you explain why and what I should do instead?

The problem is that you showed the result for a fixed $t$, but not that it holds uniformly in $t$. 
Hint: note that for each $t$ and $\varepsilon$, the inequality 
$$|X_t|\mathbb 1_{\{ |X_t|\gt\varepsilon\}} \leqslant Y\mathbb 1_{\{ Y\gt\varepsilon\}} $$
holds. Integrate and conclude by monotone convergence.
