$\sup\limits_{X \in \mathcal X, Y\in \mathcal Y}E[|Y| \mathbb 1_{\{ |X| \ge K\}}]\to0$ for $\mathcal X$, $\mathcal Y$ uniformly integrable I want to show that  $\lim_{k \to \infty } \sup_{X \in \mathcal{X}, Y\in \mathcal{Y}}E[|Y| \mathbb{1}_{\{ |X| \geq K\}}]=0$
It is given that the set of random variables $\mathcal{X}$ and $\mathcal{Y}$ is uniformly integrable
My attempt:
Since $\mathcal{Y}$ is U.I $\implies$ $\forall \epsilon >0$ , there exists $\delta >0$ such that for any $A \in \mathcal{F} $ we have 
$\mathbb{P}[A] \leq \delta \implies \sup_{X \in \mathcal{X}} E[|Y| \mathbb{1}_{A}] \leq \epsilon $
Now if we put $A=\{|X| \geq K\}$ and since X is U.I we can make $\mathbb{P}(A)$ as small as possible and therefore we get that 
$\lim_{K \to \infty} P(\{|X| \geq K\})=0$ since $X< \infty $ P.a.s and if we take $\epsilon \to 0 $ the result follow.
Does this make sense? Or is my reasoning incorrect?
I am using this to prove that $\{|X+Y| \vert X \in \mathcal{X} \text{ and } Y \in \mathcal{Y} \}$ is uniformly integrable 

Yes, it makes sense, but careful. Your set $A$ depends on $X$, but its measure can be bounded independently of $X$ using Markov's inequality and the fact that $\mathcal X$ is bounded in $\mathbb L^1$. So you fix $\varepsilon>0$, and choose a $K$ such that for each $X\in\mathcal X$, $\mathbb P\{|X| \geqslant K\}$ is smaller than $\delta$, where $\delta$ satisfies 
$$\mathbb{P}[A] \lt \delta \implies \sup_{X \in \mathcal{X}}\mathbb E[|Y| \mathbb{1}_{A}] \lt \epsilon.$$
