Suppose $0<a<1$. I’m trying to prove $\inf\{a^n:n\in\mathbb{Z}_+\}=0.$ There’s just one step in my proof I would like confirmation on.
First we note $0$ is clearly a lower bound for the set $\{a^n:n\in\mathbb{Z}_+\} $ since $a>0\implies$ by induction that $a^n>0$ for all $n\in\mathbb{Z}_+$. If $t>0$ is any other lower bound for the set $\{a^n:n\in\mathbb{Z}_+\}$, then since $a\neq0$ define $h=(1-a)/a=(1/a)-1$ so that $h>0$. Then again by induction (or just invoking the Bernoulli Inequality) one has $$(1/a)^n=1/a^n=(1+h)^n\geq1+nh$$
which is equivalent to $$a^n\leq\frac{1}{1+nh}<\frac{1}{nh}$$ since $a^n>0,1+nh>0$ for all $n$. Thus multiplying through by $h>0$ we obtain for all $n$: $$0<th\leq ha^n\leq\frac{h}{1+nh}<\frac{1}{n}$$
Now if we suppose that we already know that $\inf\{1/n:n\in\mathbb{Z}_+\}=0$, then the last inequality contradicts that fact since $th$ is a lower bound for $1/n$ greater than $0$. Thus no such $t>0$ exists, and we conclude $0$ is the greatest lower bound for the set $\{a^n:n\in\mathbb{Z}_+\} $.
My concern is with the derivation of the contradiction with $th$ and the use of $\inf\{1/n:n\in\mathbb{Z}_+\}=0$. Is this argument correct?
Your argument looks correct to me. However, you can simplify the proof.
Notice first that the sequence $\{a^n\}$ is strictly decreasing. Let $t$ be the lower bound of $S=\{a^n:n\in\mathbb{Z}_+\}$. If $t/a \gt t$, it exists $m \in \mathbb{Z}_+$ such that $t \le a^m \lt t/a$. But then $a^{m+1} \lt t$. A contradiction. Therefore $t/a = t$, which means $t=0$.