I am given the following proof of this result, but part of it is completely unclear to me:
Suppose $M$ is a nonnegative, progressively measurable, right continuous martingale. Then $P(\text{sup}_{s \leq t} M_s \ge c) \leq E(M_0)/c$
Proof:
Define $M_t^* = \text{sup}_{s \leq t}M_s$ and $\tau_c \equiv \text{inf}\{t: M_t^* \ge c\}$
Then clearly we have $\{\tau_c \leq t\} = \{M_t^* \ge c\}$
$(**)$ Now the proof claims that $M_{\tau_c} \ge c$ on the set $\{\tau_c \leq t\}$. I do not understand this.
From this point, the proof proceeds easily, as assuming $(**)$ is true we have $$P(M_t^* \ge c) = P(\tau_c \leq t) \leq \frac{1}{c}E(M_{\tau_c \wedge t} 1(\tau_c \leq t)) \leq \frac{1}{c}E(M_{\tau_c \wedge t}) = \frac{E(M_0)}{c}$$
since $\tau_c \wedge t$ is a bounded stopping time.
The reason I don't understand why $(**)$ is true is because I see clearly how $M_s^* < c$ for all $s < \tau_c$ just from the definition of $\tau_c$. I can show that $M_t^*$ is right continuous, so that (assuming $\tau_c(\omega)$ is finite) we have $M_{\tau_c}^* \ge c$. However, I do not believe that this necessarily implies $M_{\tau_c} \ge c$, because I cannot produce a contradiction from assuming $M_{\tau_c} < c$.
Any help or alternative proof strategies would be massively appreciated because I don't at all see how this works. Please help!