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Consider $$\tau_n = \frac{[2^n\tau] + 1}{2^n}$$ Note that $\tau_n$ is discrete and $\tau_n \downarrow \tau$. Now use the result you have for the discrete cases $$P(X_{\tau_n}>c\vert \tau_n=j)=\dfrac{P(X_{\tau_n}>c, \tau_n=j)}{P(\tau_n=j)}=\dfrac{P(X_{j}>c, \tau_n=j)}{P(\tau_n=j)}=P(X_{j}>c). \quad j \in \Bbb{Z}_+/ 2^n$$ Now to the continous ...
The reason continuity is relevant in showing $\{ \tau \leq t\} = \{M_t \geq b\}$ is because if $B_t$ were not continuous, then it could be that $B_t$ jumps over $b$ without hitting it, in which case $\tau$ might not occur even if $M_t > b$. As for $\{\tau \leq t\}$, this set is actually measurable with respect to \$\mathcal{F}_{\tau \wedge t} = ...