Let $(X_n)$ be a martingale with respect to $(\mathcal{F}_n)$ and suppose $\tau_1$ and $\tau_2$ are bounded stopping times such that $\tau_1\leq \tau_2<B<\infty.$ Then $$\mathbf{E}(X_{\tau_2}|\mathcal{F}_{\tau_1})=X_{\tau_1}.$$
My attempt: One may assume that $X_n$ is nonnegative. The general case will follow directly. From this link, \begin{align*}\mathbf{E}(X_{\tau_2}|\mathcal{F}_{\tau_1})&=\sum_{n\in\mathbb{N}}\mathbf{E}(X_{\tau_2}|\mathcal{F}_n)\mathbf{1}_{\{\tau_1=n\}}=\sum_{n\in\mathbb{N}}\sum_{m=n}^\infty\mathbf{E}(X_m|\mathcal{F}_n)\mathbf{1}_{\{\tau_1=n,\tau_2=m\}}\\&=\sum_{m\in\mathbb{N}}\sum_{n=1}^m\mathbf{E}(X_m|\mathcal{F}_n)\mathbf{1}_{\{\tau_1=n,\tau_2=m\}} \end{align*}
How should I proceed further here to obtain $X_{\tau_1}$?
Thanks in advance!
As pointed out by saz, using the fact that $\left(X_n\right)_{n\geqslant 1}$ is a martingale, we derive that for all $m\geqslant n\geqslant 1$, $\mathbb E\left[X_m\mid\mathcal F_n\right]=X_n$ hence continuing the computation, we get $$ \mathbf{E}(X_{\tau_2}|\mathcal{F}_{\tau_1}) =\sum_{m\in\mathbb{N}}\sum_{n=1}^mX_n\mathbf{1}_{\{\tau_1=n,\tau_2=m\}} =\sum_{n=1}^{+\infty}X_n\sum_{m=n}^{+\infty}\mathbf{1}_{\{\tau_1=n,\tau_2=m\}} . $$ Since $$ \sum_{m=n}^{+\infty}\mathbf{1}_{\{\tau_1=n,\tau_2=m\}}=\mathbf{1}_{\{n=\tau_1\leqslant \tau_2\}} $$ and since $\tau_1\leqslant \tau_2$ almost surely, the last indicator is simply $\mathbf{1}_{\{n=\tau_1\}}$ and we use an other time the result of the linked thread to get the result.