Let $A=(a_n)$ be a strictly positive sequence converge to non-zero $L$ then show that $\inf A>0$.
My attempt but I don't know how precise is my proof.
Assume for a contradiction that $\inf A<0$ then since $0<a_n$ for all $a_n$, this is a contradiction as $o$ is a lower bounded but $\inf A<0$, which contradicts the definition of being an infimum.
Assume that $\inf A=0$, then $0<a_m<\epsilon$ for some $a_m$. By the definition of convergent to $L$, we have $\forall \epsilon >0, \exists N$ such that $\forall n>N, |a_n-L|<\epsilon$. Since $\epsilon$ is arbitrary and $L$ non-zero, we assume that $m<N$. Define $a_p:=\min (a_n)$ for all $n$ between $0$ and $N$ then clearly $a_p$ is another lower bound that is greater than $0$ and hence the statement is proven.
Best Answer
You cannot say "then $0<a_m<\epsilon$ for some $a_m$" when this $\epsilon$ has not been introduced somehow before. Note that a line later you write $\forall\epsilon>0$, etc.
When the sequence $(a_n)_{n\geq1}$ with $a_n>0$ converges to some $L\ne0$ then $L>0$. Letting $\epsilon:={L\over2}$ we can say that there is an $N$ with $a_n>{L\over2}$ for all $n>N$. The finite set $\bigl\{a_n\bigm| 1\leq n\leq N\bigr\}$ consists of positive numbers; hence it contains a minimal element $a_p>0$. The number $\iota:=\min\bigl\{a_p,{L\over2}\bigr\}>0$ is a lower bound of the $a_n$ $\,(n\geq1)$ , hence $\inf_{n\geq1} a_n>0$.