Uniform Continuity: i'm right? Prove that $f : A\subset \mathbb{R} \to \mathbb{R}$ is uniformly continuous
in $A$ $ \iff$ for all
the sequences $(x_n), (y_n)\in A$ such that
$$\lim_{n\to +\infty} ( x_n - y_n )= 0$$
then 
$$\lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0.$$
$$Proof$$
Let's start with the first to demonstrate the implication:
$"{\Rightarrow}"$ 
in this case the hp are:
$ f $ uniformly continuous and $ (x_n), (y_n) $  sequences of numbers from the set $ A $ such that:
$$\lim_{n\to+\infty}(x_n - y_n) = 0.$$
and we must show that:
$$ \lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0.$$
Well, for the uniform continuity we have
$$\forall\varepsilon> 0, \exists\delta> 0 \, \: \, \ \forall x, y \in A \ | x-y | <\delta \Rightarrow | f (x)-f (y) | <\varepsilon,$$
and by definition of limit:
$$ \forall \varepsilon> 0, \exists \nu> 0 \, \, \, \: \, \, \, \ | (x_n-y_n) - 0 | = | x_n-y_n | <\varepsilon, \quad n> \nu $$
but then the fact that, thanks to the uniform continuity we have,
$$ | x-y | <\delta \Rightarrow | f (x)-f (y) | <\varepsilon, $$ 
we have that
$$ | f (x_n) - f (y_n) | <\varepsilon, \forall n> \nu $$
which is equivalent to saying
$$ \lim_ {n \to + \infty} (f (x_n) - f (y_n)) = 0. $$
$ "{\Leftarrow}" $
in this case  hp are:
$$ (1)  \lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0 $$ 
$$ (2)  \lim_{n\to+\infty}(x_n - y_n) = 0.$$ 
and we must show that $f$ is uniformly continuous:
but if   (1) and (2) are verified,  $f$ is necessarily uniformly continuous . QED
 A: Your forward implication is a bit off. In particular, you want $\delta$, not $\epsilon$, in the fourth displayed equation of your proof.
You also need to explicitly show that for any $\epsilon>0$,  there is a $N$ (your $\nu$)  that "works".  You also need to explicitly fix your $\delta$. 
You fix a value of $\epsilon$ first, and then state (and prove) that the appropriate $\delta$ and $N$ exist.   Then you verify that the choices "work".
So, you could argue as follows:
Let $\epsilon>0$. Then, since $f$ is uniformly continuous on $A$, there is a $\delta>0$ (which we are now specifying) such that 
$$\tag{1}|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon \quad\text{ for all  }x,y \text{ in } A .$$
Now, choose $N$ (as we may) so that 
$$\tag{2}|x_n-y_n|<\delta\quad\text{ for all }n>N.$$
We have found our $N$. Now we need to show that it actually works:
Then, if $n>N$, we have, by (2), that $|x_n-y_n|<\delta$. Then, from (1), we have
$|f(x_n)-f(y_n)|<\epsilon$.
So we have shown that for any $\epsilon>0$, there is an $N$ so that $n>N$ implies $|f(x_n)-f(y_n)|<\epsilon$. Thus $(f(x_n)-f(y_n))\rightarrow 0$, as desired.

For the reverse implication, it seems that you are assuming your result...
Some hints:
Prove the contrapositive statement instead.
To do this, you could do the following:
Suppose that $f$ is not uniformly continuous on $A$. Then there would be an $\epsilon>0$
such that for any positive integer $N$, you can find points $x_n$ and $y_n$ with $|x_n-y_n|<{1\over N}$ but
$|f(x_n) -f(y_n)|>\epsilon$.
