How to improve this proof about real numbers? I've trying to prove the following: let $a \in \mathbb{R}$ with $a \geq 0$ if for every real $\epsilon > 0$ we have $0 \leq a < \epsilon$ then $a =0$.
I'm sure that there is a proof much simpler the one I gave using infimum and supremum, however, I couldn't find it. I've proved it like this:
Consider the intervals $[0,\epsilon)$ for every $\epsilon > 0$, we want to show that the only $a$ in all of them is the zero number. In other words we want to prove that the intersection of all of them is just $0$. To do this we are going to show that the complement is $\mathbb{R}\setminus \{0\}$. We consider the complement:
$$\mathbb{R}\setminus\bigcap_{\epsilon \in \mathbb{R}^+}[0,\epsilon)=\bigcup_{\epsilon \in \mathbb{R}^+}(-\infty,0)\cup [\epsilon,+\infty)=(-\infty,0)\cup \bigcup_{\epsilon \in \mathbb{R}^+}[\epsilon,+\infty)$$
Now since the union is over every positive $\epsilon$ that union is of course the entire set of positive reals $(0,+\infty)$, so that the complement is $(-\infty,0)\cup(0,+\infty) = \mathbb{R}\setminus \{0\}$ and so the only element in the intersection is $0$ and hence $ a =0$.
Now is this proof good or there is a better one using $\sup$ and $\inf$? In that case, I just want a little hint on how to start with it.
Thanks very much in advance!

Proof by contradiction if $a\neq 0$ then $a>0$ so with $\epsilon =a$ we have $\epsilon\leq a$ which's a contradiction so $a=0$.
