Let $\{X_t\}_t$ be a uniformly integrable martingale with continuous path and $X_0=0$ w.r.t. filtration $\{\mathcal{F}_t\}_t$. Denote $X_\infty = \lim_{t\to\infty}X_t$. Let $T$ be any stopping time. Prove that if $\mathbb{E}[|Y_T-Y_\infty|]\leq C\cdot P(T<\infty)$ for some constant $C$, then $\mathbb{E}[|Y_T-Y_\infty|\,|\,\mathcal{F}_T]\leq C$.
Let $I_F$ be the indicator function of any set $F\in\mathcal{F}_T$. We have $$\mathbb{E}[\mathbb{E}[|Y_T-Y_\infty|\,|\,\mathcal{F}_T]I_F] = \mathbb{E}[|Y_T-Y_\infty|I_F] \leq C\cdot P(F)\tag{$*$}$$ The next step is to conclude $\mathbb{E}[|Y_T-Y_\infty|\,|\,\mathcal{F}_T]\leq C$. If $(*)$ holds, then I know how to get the desired property now, but the real difficulty is how to get the inequality in $(*)$.
I finally work out this question following Kavi's advice. I post my solution for other people's further interest.
To show $(*)$ holds, let $F\in\mathcal{F}_T$, and define similarly \begin{equation*} T^F(\omega) = \begin{cases} T(\omega) & \text{if $\omega\in F$} \\ \infty & \text{otherwise} \end{cases} \end{equation*} Then $T^F\geq T$ and $T^F$ is a stopping time because $\{T^F\leq t\}=F\cap\{T\leq t\}\in\mathcal{F}_t$ for $t\geq 0$. By assumption, \begin{align*} \mathbb{E}(|Y_\infty-Y_{T^F}|^pI_F)+\mathbb{E}(|Y_\infty-Y_{T^F}|^pI_{F^c}) &= \mathbb{E}(|Y_\infty-Y_{T^F}|^p) \\ &\leq C\cdot P(T^F<\infty) \\ &= C\cdot P(F) \end{align*} Notice that $\mathbb{E}(|Y_\infty-Y_{T^F}|^pI_{F^c})=0$, and $\mathbb{E}(|Y_\infty-Y_{T^F}|^pI_F)=\mathbb{E}(|Y_\infty-Y_T|^pI_F)$. Thus, for any stopping time $T$ and $F\in\mathcal{F}_T$, we have $\mathbb{E}(|Y_\infty-Y_T|^pI_F)\leq C\cdot P(F)$.
Once $(*)$ holds, then let $F=\{\mathbb{E}(|Y_\infty-Y_T|^p\,|\,\mathcal{F}_T)>C\}$ and suppose $P(F)>0$, \begin{equation*} \mathbb{E}(\mathbb{E}(|Y_\infty-Y_T|^p\,|\,\mathcal{F}_T)I_F) > C\cdot P(F) \end{equation*} which gives a contradiction. Thus, $P(F)=0$ and $\mathbb{E}(|Y_\infty-Y_T|^p\,|\,\mathcal{F}_T)\leq C$ almost surely.