How do I Show that $\exists a, \forall e > 0; a < e \Rightarrow a \le 0$

When proving using contradiction,can the method specified below be used.

lets assume that $\forall a, \exists e > 0; a < e$ and $a > 0$

since the condition holds for all 'a', let $a=2e$then $2e<e$, $2<1$. Which is a contradiction.therefore $\exists a, \forall e > 0; a < e \Rightarrow a \le 0$.

Can I know where i might have gone wrong and if not, whether theres another method to prove it.