Let $s < t$. We have to show that $E[M_t^\tau \mid \mathcal{F}_s] = M_s^\tau$ almost surely.
For each $n$ we know, as you mentioned, that $(M^\tau)^{\sigma_n} = M^{\tau \wedge \sigma_n}$ is a martingale, where $\sigma_n$ is the localizing sequence of stopping times. Thus $E[M^{\tau \wedge \sigma_n}_t \mid \mathcal{F}_s] = M^{\tau \wedge \sigma_n}_s$ a.s. (*)
(If you are worried about the boundedness of stopping times here, you can assume without loss of generality that each $\sigma_n$ is bounded, by replacing it with $\sigma_n \wedge n$.)
Now $M_s^{\tau \wedge \sigma_n} = M_{s \wedge \tau \wedge \sigma_n}$. By assumption $\sigma_n \uparrow \infty$ a.s., so $M_s^{\tau \wedge \sigma_n} \to M_s^\tau$ almost surely. The same holds with $t$ in place of $s$. Moreover, $|M_s^{\tau \wedge \sigma_n}| \le \varepsilon$ a.s. for all $n$.
By the conditional dominated convergence theorem, with dominating function $\varepsilon$, we can conclude $E[M^{\tau \wedge \sigma_n}_t \mid \mathcal{F}_s] \to E[M^{\tau}_t \mid \mathcal{F}_s]$ a.s. So we pass to the a.s. limit on both sides of (*) and get the result.
Note that this same argument shows the following version of the optional stopping theorem, which is really good to know:
Let $M_t$ be a continuous martingale and $\tau$ a (not necessarily bounded!) stopping time. Suppose that the process $M_{t \wedge \tau}$ is bounded, i.e. there is a constant $C$ such that $|M_{t \wedge \tau}| \le C$ a.s. for all $t$. (You might say: "$M$ is bounded up to time $\tau$.") Then $E[M_\tau] = E[M_0]$.
This is the version that Evan Aad was referring to in the question I linked in my comment. More generally, it also holds if $\{M_{t \wedge \tau} : t \ge 0\}$ is uniformly integrable. As a general principle for martingales, anything that's true for bounded stopping times $\tau$ will also be true if the process is bounded up to $\tau$.
