Proof verification : Show that $\sup \left\{3-\frac{2}{n^2}:n\in\mathbb{N}\right\} =3$ 
Show that $\sup \left\{3-\frac{2}{n^2}:n\in\mathbb{N}\right\} =3$

I just want to verify my proof (assuming I have already proven that $\forall x\in A, x\lt3$
)
$$A :=\left\{3-\frac{2}{n^2}:n\in\mathbb{N}\right\} $$

proof(contd.)
  $$\forall\epsilon>0,\exists x_\epsilon\in A \text{ s.t. } x_\epsilon > 3-\epsilon$$
  $$\text{ Therefore, } 3-\frac{2}{n^2}>3-\epsilon\\
\text{ for } n\in\mathbb{N}\text{, } n>\sqrt{\frac{2}{\epsilon}}
$$
  $$\text{ Thus, the above enequality is true for all}\\
n = \left[\sqrt{\frac{2}{\epsilon}}\text{  }\right]+1\text{  }(\text{integer part plus one})\text{ Q.E.D.}$$


Are there any drawbacks in my logic?
 A: As Jack already mentioned there are some parts you need to fix.
In general the best way is to try to guess what the supremum or infimum of the set is and to proofe it by contradiction:
Assume $3$ is not the supremum. This means there exists $a>0$ such that $s:=3-a$ is the supremum. But since $1/n^2 \to 0$ as $n\to \infty$ you can find an index $N_a$ such that $1/n^2 <a/2$ for all $n\geq N_a$ and hence we have $s<x_n$, where $x_n:=3-1/n^2$ and $n\geq N_a$. This is impossible and hence $3$ is the supremum of the set.
A: 
I just want to verify my proof (assuming I have already proven that $\forall x\in A, x\lt3$
  )
  $$
A =\left\{3-\frac{2}{n^2}:n\in\mathbb{N}\right\} 
$$
  proof(contd.)
  $$
\forall\epsilon>0,\exists x_\epsilon\in A \text{ s.t. } x_\epsilon > 3-\epsilon
$$
  (True, but why? This is basically what you are trying to show in the second step. Once you have this, together with the "assuming I have proven that" statement, you are done.)
Therefore,(Hardly can I understand what this "Therefore" means. It seems that you next sentence has nothing to do with the previous arguments.) 
  $$
3-\frac{2}{n^2}>3-\epsilon
$$
  $$
\text{ for } n\in\mathbb{N}\text{, } n>\sqrt{\frac{2}{\epsilon}}
$$
  Thus, the above enequality (Which "inequality" are you referring to? This "Thus" is very confusing.) is true for all
  $$
n = \left[\sqrt{\frac{2}{\epsilon}}\text{  }\right]+1\text{  }(\text{integer part plus one})\text{ Q.E.D.}$$
  Are there any drawbacks in my logic? (Yes. See comments above.)


To fix the problem, you would need to show (instead of claiming) that for given $\epsilon>0$, there exists $x\in A$ such that $x>3-\epsilon$. Using the definition of $A$, it means that there exists $n\in{\bf N}$ such that
$$
3-\frac{2}{n^2}>3-\epsilon.
$$
I think you have already known how to do this. 
