Why does this proof of limit of a sequence: $\lim_{n\rightarrow \infty}\frac{1}{n}\neq0$ fail? Here is a proof of:
$$\lim_{n\rightarrow \infty}\frac{1}{n}\neq0, n \in \mathbb{N}$$
As per definition of a limit of a sequence:
$$ \lim_{n\to\infty}S_n=A\iff\forall \epsilon\in\Bbb R_{>0}: \exists N\in\Bbb N : \forall n\in \Bbb N_{>N} : |(S_n-A)|<\epsilon,$$ 
And the proof is simple, since $\frac{1}{n}$ is a rational number and $\epsilon$ is a real number, there always exists a real number between 0 and any rational number. Because of that there exists an $\epsilon$ for which the condition can not hold.
Why is this proof wrong?

You are confused by the order of the quantifiers. If you take a look at the formula you listed, it says that given any positive number there is some positive integer such that every term of the sequence with index no less than the positive integer is close to the "limit" by at most the positive number. So the upper bound $\varepsilon$ is given first, not the other way around.
