Help: Real Analysis Proof: Prove $|x| < \epsilon$ for all $\epsilon > 0$ iff $x = 0$. I've recently started studying Real Analysis by myself(not going to any school at the moment) and need some help to review my answer to the question below.
Question
Let $x \in \mathbb{R}$. Prove $|x| < \epsilon$ for all $\epsilon > 0$ iff $x = 0$.
Answer
Suppose for contradiction that $x \in \mathbb{R}$, $|x| \geq \epsilon$ for all $\epsilon > 0$.
$|x| = |x - \epsilon + \epsilon| \leq |x - \epsilon| + |\epsilon|$ by Triangle inequality.
Since the first term($|x - \epsilon| \geq 0$) and by our assumption($\forall \epsilon > 0$), this contradicts $|x| \geq \epsilon$.
Comment
I am not sure if my answer is solid enough to lead to the contradiction. Any help or comment is appreciated.
Source
The question is taken from the Exercise 2.12 of the pdf below.


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*http://www.math.louisville.edu/~lee/RealAnalysis/
Edit
Thank you for all the questions!!
It seems like I've stepped towards the wrong direction from the beginning!
I will try myself to prove that there is an arbitrary $\epsilon$ which can take the value below $|x|$ unless $|x| = 0$ based on $\forall \epsilon > 0$
 A: If $x=0$ then certainly, $|x|=|0|=0<\epsilon$ for every $\epsilon>0$.
Now, you want to show that if for all $\epsilon>0$ , $|x|<\epsilon$ then, $x=0$.
Suppose $x\neq 0$ so $|x|>0$. Hence, for $\epsilon=|x|$, $|x|<|x|$. Contradiction.
A: You want to prove that
$$(\forall x\in \Bbb R)\;\;\Bigl( (\forall \epsilon>0)\; \; |x|<\epsilon\;\; \iff \;\; x=0\Bigr)$$
Proof
Let $ x\in \Bbb R$ such that
$$(\forall \epsilon>0) \;\; |x|<\epsilon$$
Assume $ x\ne 0$.
$$x\ne 0 \implies |x|>0$$
$$\implies \;\; \exists \epsilon(=\frac{|x|}{2})>0 \;\; : \;\; |x|>\epsilon $$
which contradicts the hypothesis.
A: If $x = 0$, the result follows. Suppose otherwise that $|x| < \varepsilon$ for every $\varepsilon > 0$ and $x\neq 0$. Then either $x > 0$ or $x < 0$. In the first case $(x > 0)$, one can choose $\varepsilon = x/2$, whence we get
\begin{align*}
|x| = x < \varepsilon = \frac{x}{2} \Longleftrightarrow x < 0 
\end{align*}
which leads to a contradiction. Similarly, if $x < 0$, we can choose $\varepsilon = -x/2$, whence we get that
\begin{align*}
|x| = -x < \varepsilon =  -\frac{x}{2} \Longleftrightarrow x > 0
\end{align*}
which is also a contradiction. Consequently, $x = 0$ and we are done.
A: Assume that $x \neq 0$. Then, $|x| = c >0$. Set $\epsilon <c$ for example, $\epsilon = \frac{c}{2}$. Then we have $c = |x|  < \epsilon = c/2$ by assumption, meaning that $c < c/2$. But this of course cannot happen.
