= Solution
Fix $x>0$ and let $A=\{M_0<x\}\in\mathcal F_0$. On $A$, <continuity> of the paths implies $0\leq M_{t\wedge T_x}\leq x$, and $M_{T_x}=x$ on $\{T_x<\infty\}$. Bounded-time <optional stopping> gives, for every $t$,
$$
\mathbb E[\mathbf1_A M_{t\wedge T_x}\mid\mathcal F_0]=\mathbf1_A M_0.
$$
On $A\cap\{T_x=\infty\}$, the stopped process is $M_t\to0$; on $A\cap\{T_x<\infty\}$ it eventually equals $x$. The bound by $x$ permits conditional <dominated convergence>, yielding
$$
x\mathbf1_A\mathbb P(T_x<\infty\mid\mathcal F_0)=\mathbf1_A M_0.
$$
On $A^c$, $T_x=0$, so the conditional hitting probability is one. Finally, a continuous nonnegative path tending to zero attains any positive supremum: after a sufficiently late time it is below half that supremum, and its maximum on the earlier compact interval is attained. Thus $\{M^*\geq x\}=\{T_x<\infty\}$, including equality at the level. We have proved the conditional form of the <maximal identity for a continuous nonnegative local martingale tending to zero>:
$$
\boxed{\mathbb P(M^*\geq x\mid\mathcal F_0)=1\wedge\frac{M_0}{x}.}
$$
The restriction to $A$ before taking the limit is important: $M_0$ need not be bounded, and the unstopped <martingale> need not have <uniform integrability>.
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