Would someone please give me some help with my case 2 below?

The problem: Let x $\displaystyle \ge $ 0, x $\displaystyle \in $ $\displaystyle \Re $ such that for any $\displaystyle \varepsilon $ > 0, x $\displaystyle \le $ $\displaystyle \varepsilon $. Prove that x = 0

Solution:

Case 1: Let x > 0 and $\displaystyle \varepsilon $ = $\displaystyle \frac{x}{2}$

then

$\displaystyle \varepsilon $ = $\displaystyle \frac{x}{2}$ > 0 because x > 0

since $\displaystyle \varepsilon $ > 0 by given, x $\displaystyle \le $ $\displaystyle \varepsilon $

Case 2: Let x = 0 and $\displaystyle \varepsilon $ = $\displaystyle \frac{x}{2}$

then

$\displaystyle \varepsilon $ = 0/2 = 0

since $\displaystyle \varepsilon $ = 0 = x, I cannot conclude x $\displaystyle \le $ $\displaystyle \varepsilon $

unless I let $\displaystyle \varepsilon $ = x + y (where y $\displaystyle \in $ $\displaystyle {\Re ^ + }$). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.

Thanks,