If $T=\left\{\frac{1}{n}\ |\ n\in\mathbb{N}\right\}$ show that $\inf T=0$.
I have seen proofs of this question using Archimedean property, but was wondering if it was enough to show that a lower bound can't exist if we let $L$ denote lower bound of $T$ such that $L>0$.
To prove that $L>0$ cannot be the infimum, you need to show that no such $L$ is a lowerbound. This means, you need to argue that there exists some $n$ such that $\frac{1}{n} <L$, or equivalently that $n >\frac{1}{L}$. This is exactly the Archimedian property.
So, if you write down the details of your approach, you are going to run into the Archimedian property.