$\{X_n, n\geq 1\}$ is a renewal process. Let, there exists an $\alpha>0$ such that $P(X_n\geq \alpha)>0$. Now, define another related renewal process, $\{\bar X_n, n\geq 1\}$ by, \begin{equation} \bar X_n= \begin{cases} 0, & \text{if}\ X_n <\alpha \\ \alpha, & \text{otherwise} \end{cases} \end{equation} For, $n\ge 1$, let $S_n = X_1+\dots + X_n$ and $\bar S_n = \bar X_1+\dots + \bar X_n$. Also, let, $N(t) = \max \{n\mid S_n \leq t\}$ and $\bar N(t) = \max \{n\mid \bar S_n \leq t\}$
Show that, $\bar N(t) \geq N(t)$.
What I did: It is easy to see that $\bar X_n \leq X_n$, so $\bar S_n\leq S_n$. We also know that $\{N(t)\geq n\}\Leftrightarrow\{S_n\leq t\}$.
The braces $\{$ and $\}$ in your last statement should not be there. Then we get:$$N(t)\geq n\iff S_n\leq t$$as it should.
Combining this with $\overline S_n\leq S_n$ we find:$$N(t)\geq n\implies \overline S_n\leq t$$or - on base of $\overline N(t)\geq n\iff \overline S_n\leq t$ - equivalently:$$N(t)\geq n\implies \overline N(t)\geq n\tag1$$
This for every $n$, so it implies that: $$\overline N(t)\geq N(t)\tag2$$
Formally pick out an arbitrary $\omega\in\Omega$ an let it be that $n=N(t)(\omega)$. Then $N(t)(\omega)\geq n$ so according to $(1)$ we have $\overline N(t)(\omega)\geq n=N(t)(\omega)$. This proves that $\overline N(t)(\omega)\geq N(t)(\omega)$ for every $\omega\in\Omega$ or shortly that $(2)$ is valid.