Absolute value zero is always less than a positive number I'm trying to prove that $$a=0 \iff \vert a\vert<\epsilon,\forall \epsilon>0$$
First, lets prove that $a=0 \rightarrow\vert a\vert<\epsilon,\forall\epsilon>0$. This is proved by contradiction, we assume that $$a=0$$and there exists an $\epsilon >0$, such that $$\vert a\vert\geq\epsilon$$ Since $a=0$, then $\vert a \vert = 0$. In other words, there would have to be an $\epsilon>0$, such that $0\geq \epsilon$, which clearly can't be true. This proves that left implies right. Next, we must show that $$\vert a \vert<\epsilon \to a=0$$
For all $\epsilon >0$. My idea was to interpret $\vert a \vert < \epsilon$ as a set of elements, for which the distance from zero is always less than $\epsilon >0$, which means the set has only a single element, zero. How can I show this rigorously?
 A: For the first part, you don't need proof by contradiction, and indeed, it only makes it more complicated to see what's really happening. It's quite simple.

$a=0$ implies that $|a|=0$, so the sentence $\forall \epsilon > 0: |a| < \epsilon$ is equivalent to the sentence $\forall \epsilon > 0: 0 < \epsilon$, which is trivially true, since the statements $\epsilon > 0$ and $0<\epsilon$ are identical.


For the second part, first of all, a warning, your sentence:

Next, we must show that $$\vert a \vert<\epsilon \to a=0$$
For all $\epsilon >0$.

is quite slopy, and read as is, untrue. What you must show is that $$(\forall \epsilon > 0: |a|<\epsilon) \rightarrow (a=0).$$ The scope of the quantifier is very important here.
To actually solve the second part, I suggest you either use proof by contradiction, or even better, contraposition. That is, instead of proving the statement above, prove the equivalent statement
$$(a\neq 0)\rightarrow (\exists \epsilon > 0: |a|\geq \epsilon)$$
(remember, the statements $A\rightarrow B$ and $\neg B\rightarrow \neg A$ are equivalent).
A: The idea you present for proving the reverse direction will probably not be fruitful since the set you describe has in fact infinitely many elements for any $\epsilon > 0$.
The most natural idea here is to try a proof by contradiction as others have pointed out. Another idea is to consider a function $f(x)$ sandwiched between $1/x$ and $-1/x$, like $-1/x<f(x)<1/x$. Then it follows that $\lim_{x\rightarrow \infty}f(x) = 0$. So if we now show that $\lim_{x\rightarrow \infty}f(x) = a$ we are done. I.e. we wish to show
$$\forall \epsilon > 0 \hspace{2mm} \exists x_0 \in \mathbb{R} \hspace{2mm} s.t. \forall x > x_0 \hspace{2mm} \left | f(x) - a \right| < \epsilon$$
Now, 
$$\left | f(x) - a \right| \leq \left | f(x) \right | + \left |a \right | \\ < \left | 1/x \right | + \left | a \right | $$
So we can just take $x_0 = \frac{1}{\epsilon - \left |a \right |}$. This is always a positive well defined number by the antecedent.
A: Again prove by contradiction. Suppose $a \neq 0$ and $|a| <\epsilon$ for every $\epsilon >0$.  Let $\epsilon =\frac 1 2 |a|$. This is a positive number since $|a| >0$. Thus we get $|a|<\frac 1 2 |a|$. Do you see a contradiction  here? 
A: 
Next, we must show that $$\vert a \vert<\epsilon \to a=0$$
  For all $\epsilon >0$. 

No, you do not need to show that. It’s not true. 
What you wrote says literally that if you take any positive number $\epsilon$, for example, $\epsilon=2,$ the only  number $a$ whose absolute value is less is zero. But there are infinitely many numbers with absolute values less than $\epsilon.$
What you wrote is equivalent to
$$\forall \epsilon>0.( a=0 \iff \lvert a\rvert<\epsilon ),$$ 
with parentheses to show what applies to what.
Here’s the original statement to be proved:
$$a=0 \iff (\forall \epsilon>0.( \lvert a\rvert<\epsilon )).$$
For the $\impliedby$ direction, you first assume that $\lvert a\rvert<\epsilon $ for every positive number $\epsilon $.
You can do this by contradiction: make a second assumption, that $a\neq0,$ and show that the first assumption cannot also be true. 
