Set of Non-Negative Real Numbers is not Well-Ordered by Usual Ordering

Theorem
The set of non-negative real numbers $\R_{\ge 0}$ is not well-ordered under the usual ordering $\le$.

Proof
$\R_{\ge 0}$ is a well-ordered set.

Then by definition, all subsets of $\R_{\ge 0}$ has a smallest element.

Take the subset $\R_{> 0}$:
 * $\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$

Suppose $x \in \R_{> 0}$ is a smallest element.

Then $\dfrac x 2 \in \R_{> 0}$.

But $\dfrac x 2 < x$, which contradicts the supposition that $x \in \R_{> 0}$ is a smallest element.

Hence there can be no such smallest element.

So by Proof by Contradiction, $\R_{\ge 0}$ is not well-ordered by $\le$.