Proof of $a \le 0$ $\Leftarrow \Rightarrow$ $\forall \epsilon > 0$ $a < \epsilon$

Question

In analysis class I saw a proof but I would like see another

Proof:

$\Rightarrow$
Suppose that $a \le 0$ $\land$ $\epsilon > 0$

if $a=0$ by hypothesis $a < \epsilon$
if $a<0$ $\Rightarrow$ $a < \epsilon$

$\Leftarrow$
Suppose that $\epsilon > 0$ $\land$ $a < \epsilon$
$0<a$ $\land$ $a<\epsilon$ $\Rightarrow$ $0 < \epsilon$
$a<0$ $\land$ $0<\epsilon$ $\Rightarrow$ $a < \epsilon$

$0<a$ $\land$ $a<0$ is a contradiction, then by a order axiom $a=0$ $\Rightarrow$ $a \le 0$

I honestly don't know if my proof is correct, specifically $\Leftarrow$

Is correct my proof?

Answer 1

For $\Leftarrow$, suppose $a>0$ and take $\epsilon = a/2 >0$.