I'm trying to understand the difference between $0<m<f(x)\,, \forall x$ versus $0<f(x)\,, \forall x$.
My best current guess is the following:
$0<m\le f(x)\,, \forall x$ implies that there is always a gap between $f(x)$ and $0$, such that even in the limit, $f(x) \ne 0$. E.g. it would be impossible for $\lim_{x \to \infty} f(x) =0$
$0<f(x)\,, \forall x$ implies it is possible for $\lim_{x \to \infty} f(x) =0$
Is the above interpretation correct?
If so, then I'm still having difficulty understanding why, if $f(x)=m$ it is still impossible for $\lim_{x \to \infty} f(x) =0$.
You're right: it's different even if $f$ is continuous. For example, the function $f:(0,\infty)\rightarrow \mathbb{R}$ given by $f(x)=1/x$ is continuous, yet $\lim_{x\rightarrow \infty}f(x)=0$.
On the other hand, if $f:I\rightarrow\mathbb{R}$ is continuous and positive at some point $a$, then in fact $f$ is positive on some open interval containing $a$. Indeed, the set $U:=f^{-1}((0,\infty))=\{x\in I : f(x)>0\}$ is open because $f$ is continuous, and $a\in U$: thus there is some open interval $J$ such that $a\in J\subseteq U$. $$f(x)>0 \quad \text{ for all } x\in J.$$ Shrinking $J$ so that $J$ is closed+bounded and $J\subseteq U$, in fact this means $f$ is a positive function on a compact set, so it achieves its minimum. Thus there is some lower bound $M>0$ so that $f(x)\geq M>0$ for all $x\in J$.
It really all depends on which kinds of functions $f$ you are referring to in your OP.