Let $(X_n)_{n\geq0}$ be a discrete-time martingale, and let $T$ be an almost surely finite stopping time such that $$\mathbb{E}(|X_T|)<\infty,\hspace{2cm}\lim_{n\to\infty}\mathbb{E}(|X_n|1_{\{T>n\}})=0.$$ Show that the stopped process $(X_{T\wedge n})_{n\geq0}$ is a uniformly integrable martingale.
My attempt so far: For $n\geq0$ we have $$|X_{n\wedge T}|=|X_T|1_{\{T\leq n\}}+|X_n|1_{\{T>n\}}.$$ We have that $|X_T|1_{\{T\leq n\}}$ is increasing to $|X_T|$, so by monotone convergence and the assumptions on the stopping time $T$ we have $$\lim_{n\to\infty}\mathbb{E}(|X_{n\wedge T}|)=\mathbb{E}(|X_T|)<\infty.$$ Additionally, as $X^T$ is a martingale we have that $|X^T|$ is a submartingale, and so $\sup_{n\geq0}\mathbb{E}(|X_{n\wedge T}|)=\lim_{n\to\infty}\mathbb{E}(|X_{n\wedge T}|)=\mathbb{E}(|X_T|)<\infty$, (not sure how rigorous that supremum to limit argument is) and thus $X^T$ is $L^1$-bounded. I'm not sure unfortunately how to show uniform integrability, and so any help / advice would be greatly appreciated!
You wrote $\lvert X_{n\wedge T}\rvert$ as $Y_n+Z_n$, where $Y_n=|X_T|1_{\{T\leq n\}}$ and $Z_n=|X_n|1_{\{T>n\}}$. Since the sum of uniformly integrable sequences is a uniformly integrable sequence, it suffices to show that $(Y_n)_{n\geqslant 1}$ and $(Z_n)_{n\geqslant 1}$ are uniformly integrable.
For $(Y_n)_{n\geqslant 1}$, it suffices to notice that $0\leqslant Y_n\leqslant \left\lvert X_T\right\rvert$ and by assumption, $\left\lvert X_T\right\rvert$ has a finite expectation.
For $(Z_n)_{n\geqslant 1}$, the assumption gives that $\lVert Z_n\rVert_1\to 0$, which is a sufficient condition for uniform integrability, as it was shown here.