In one of my textbooks it says let $\tau = \min\{m : Y_m \ge x\}$ be the first passage time of $Y$ above the level $x$ (where $Y$ is a martingale). Then, $\tau \wedge n$ is a bounded stopping time.
However, from what I understand, $\tau \wedge n$ would have to be bounded if $\tau \wedge n \le K$ for all $n$ and some $K \in \mathbb{R}$. But in this case, $$ \tau \wedge n = \begin{cases} \tau & \text{ if }\tau < n \\ n & \text{ if }n < \tau \end{cases} $$
which is completely dependent on $n$. I'm not completely sure if I am missing something simple? But it seems to me that we would need to know $\mathbb{P}(\tau < \infty) = 1$ before making this claim.
It is not $\tau \wedge n \leq K$ for all $n$ and some $K \in \mathbb{R}$, that you want, but for each fixed $n$, $\tau(\omega) \wedge n \leq K$ for all $\omega \in \Omega$ and some $K \in \mathbb{R}$, where $\Omega$ is the sample space.
In other words, for $n \geq 1$ and a stopping time $\tau$, let $$\tau_n = \tau \wedge n$$ Now, $\tau_n$ is the minimum of two stopping times (yes, the deterministic time $n$ is itself a stopping time -- check!), hence a stopping time itself. By definition, $\tau_n \leq n$ with probability 1, hence $$\mathbb{P}(\tau_n < \infty) = 1.\tag{1}$$ The advantage is that (1) holds even if $\mathbb{P}(\tau < \infty) < 1$.