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