Cannot prove this:
If $x$ has the property that $0 \leq x < h$ for every positive real number $h$, then $x = 0$.
I tried proving this by contradiction, by setting $x > 0$. However, it lead me nowhere.
There are other answers which suggest we take $h = \frac{x}{2}$ and simply show $0 < x/2 < x$, which is contradiction, but this is plain wrong (in my opinion). $x$ denotes a set, and by choosing $h = x/2$, there still exists some $0 < x_2 < x/2$. $x$ is a notation for a set, since Apostol says "x has a property" in the exercise above^. Wiki difines it as
In mathematics, a property is any characteristic that applies to a given set
I think the suggestion solves your problem.
Put \begin{equation} A=\{ x \in \mathbb{R}; 0\leq x< h, \;\forall h>0\}. \end{equation} If $x > 0$ and $x \in A$ then $\frac{x}{2} >0$ and we can choose $h=\frac{x}{2}$. By definition we will have to have \begin{equation} 0<x<\frac{x}{2}. \end{equation} From there we have \begin{equation} x<\frac{x}{2} \; => \; 2x<x \;=>\; x<0, \end{equation} thats your contradiction.