Question: Does there exist a matrix $A$ in $\Bbb{R}^{3\times 3}$ that has only $0$ as eigenvalue? 

As far as I think, an upper diagonal matrix with only zeros on the diagonal is such an example. 

And if I'm correct, is such a matrix diagonalizable?

Answer: 
The equivalences $$\lvert A\rvert< B\iff\begin{cases}B\ge 0\\ A> -B\\ A< B\end{cases}\\ \lvert A\rvert\ge B\iff B\le 0\lor \begin{cases}B> 0\\ A\le -B\end{cases}\lor\begin{cases}B>0\\ A\ge B\end{cases}$$

are a good place to start from.
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation

$$\frac1{|x|-3}=\frac12,$$ which gives $|x|=5$. The inequation holds for $|x|\ge5$. Your resolution was right.

So the solution set is

$$(-3,3)\cup(-\infty,-5]\cup[5,\infty).$$