I used the definition of $\inf $ to prove this. Let $A=\{\frac{1}{n}: n \in \mathbb{N}\}$. Since, $0<\frac{1}{n}$ for all $ n \in \mathbb{N}$, $0$ is a lowerbound of $A$. And, by the Archimedean property, given any $\frac{1}{n} \in A$, there exists $m \in \mathbb{N}$ such that $\frac{1}{m}<\frac{1}{n}$.In other words, we have proved that any number greater than the glb is not an lower bound. Hence, we are done.
Is this rigorous enough?
On
You've got the right idea to first establish that $0$ is a lower bound for $A$, which is evident. I think a good way to proceed after this is by showing that, for any $\varepsilon>0$ we have $A\cap [0,\varepsilon)\neq\emptyset$. This will show that $0=\inf(A)$ by the characterisation of infimum. We can show that $A\cap [0,\varepsilon)\neq\emptyset$ by showing that $[0,\frac{1}{n})\neq\emptyset$ for any $n=1,2,3,\ldots$, which is fairly easy to argue.
You showed that $0$ is a lower bound, and that no element of $A$ is a lower bound of $A$. You have to show that no element of $\mathbb R$ greater than $0$ is a lower bound of $A$. You might try this: let $x\in\mathbb R$ with $x>0$. By the archimidean property, there exists $n\in\mathbb N$ with $n\geq 1/x$. Now finish the argument.