We define $\{X_t\}_{t \in [0,\tau) }$ as a continuouts local martingale on the arbitrary and possibly random time interval $[0,\tau)$.
We let $T_k$ be the "reducing sequence from above restricted to $[0,\tau)$".
- Does this mean $T_k:=\text{inf} \{t \in [0,\tau): |X_t|>k\}$?
We define the stopping time $\gamma:[0,\infty) \rightarrow [0,\tau)$ as
$ \gamma=\begin{cases}t-(k-1)&\text{if } T_{k-1}+(k-1) \leq t \leq T_k+(k-1),\\T_k&\text{if } T_k+(k-1) \leq t \leq T_k+k\end{cases}$
Now we let $n=\lfloor t \rfloor +1$ and imply
$\gamma(t) \leq min(T_n,t)$
- Why does this last equality hold?
Your definition of $T_k$ looks reasonable, although let me point out that the phrase you quoted does not have a standard definition. Also, it appears (from context clues) that $k\in\mathbb N=\{1,2,\ldots\}$, and if so it should be stated as part of the definition of $T_k$.
To show that $\gamma(t)\leq \min(T_n,t)$ is the same as showing that $\gamma(t)\leq T_n$ and $\gamma(t)\leq t$, which can be done separately of each other. First, to show that $\gamma(t)\leq t$, observe in the first case of the $\gamma$ definition that $t-(k-1)\leq t$ since $k\geq 1$, and in the second case of the $\gamma$ definition we have that $T_k+(k-1)\leq t$ and thus $T_k\leq t$. Thus $\gamma(t)\leq t$. Likewise, $\gamma(t)=T_k$ if we are in the second case of the definition, so certainly $\gamma(t)\leq T_k$ in that case. Finally, in the first case of the $\gamma$ definition, we have $t\leq T_k+(k-1)$ and $\gamma(t)=t-(k-1)$, and substituting the former inequality into the latter equation yields $\gamma(t)\leq T_k$, as desired.