Why would $\inf A = 0$ if $A \cap [- \infty, \epsilon] \not = \emptyset$, $A \subset [-\epsilon,\infty]$? Where $\epsilon > 0$.
By assumption we know that $A$ is non-empty. We have a result:
$$A \subseteq B \implies \inf B \leq \inf A$$
So $$-\epsilon \leq \inf A$$
How can I argue further?
By def. of infimum. $s$ is infimum of $A$, if for every $\epsilon >0$ there exists $a \in A$ s.t. $a < s + \epsilon$.
This could suggest:
Let $a=-\epsilon$, then assume $\inf A = s$. Then must hold $a < s + \epsilon$. However, now it's possible that $s=-\epsilon$, since $a < -\epsilon + \epsilon$ is true.