Prove that $f$ is uniformly continuous iff there exist sequences $a_n,b_n$ such that if $\lim a_n=\lim b_n \implies \lim f(a_n)=\lim f(b_n)$ 
Suppose $f$ is a real-valued function, and $a_n,b_n$ are real sequences.
Prove that $f$ is uniformly continuous $\iff (\lim a_n=\lim b_n \implies  \lim f(a_n)=\lim f(b_n))$




*

*Suppose $f$ is uniformly continuous. Then, in particular, $f$ is continuous. Suppose $a_n\to x$ as $n\to\infty$. We know by continuity of $f$ that: $$\lim a_n=x\implies \lim f(a_n)=f(x)$$ We know that $\lim a_n=\lim b_n=x$ so that also: $$\lim b_n=x\implies \lim f(b_n)=f(x)$$
Together this gives that if $f$ is uniformly continuous, that $\lim f(a_n)=\lim f(b_n)$.



*

*Suppose $f$ is not uniformly continuous. We then know that:
$$\forall_{\delta>0}\exists_{\epsilon>0}:|x-a|<\delta \not \Rightarrow |f(x)-f(a)|<\epsilon$$
Suppose we know that $\lim a_n=a=\lim b_n=b$, we want to show that $\lim f(a_n)\neq f(b_n)$. 
As the sequences are arbitrary in $\mathbb{R}$, we know that:
$a-b=0\implies|a-b|<\delta$. 
As $f$ is not uniformly continuous, we know that for a particular $\delta>0$ we pick, the following holds:
$$|a-b|<\delta \not \Rightarrow |f(a)-f(b)|<\epsilon \ \ \ (*)$$
Now suppose that $\lim f(a_n)=\lim f(b_n)$ so that $f(a)=f(b)$. Then for any $\epsilon>0$:
$$|a-b|=0<\delta\implies|f(a)-f(b)|=0<\epsilon$$
This is contradictory with the statement $(*)$; non-uniform continuity. Thus: $f(a)$ cannot equal $f(b)$ if $a=b$ so that if $f$ is not uniformly continuous, we know that: $$\lim a_n = \lim b_n \not \Rightarrow \lim f(a_n) = \lim f(b_n)$$ $\tag*{$\Box$}$ 
 A: There are various things wrong here.
In the title of your question you require the existence of two sequences; but in the body it becomes clear that you mean: for all pairs of sequences satisfying $\lim a_n=\lim b_n$ something happens. In reality the criterion outlined in skin tone just describes ordinary continuity: Take all $b_n=b$, and you obtain the standard definition.
When it comes to your proof the first big mistake is your description of "not uniformly continuous". It does not mean that for all $\delta>0$ there is an $\epsilon>0$ such that something disagreable happens. Instead it means that there is an $\epsilon_0>0$ such for all $\delta>0$ you can find two points $x$, $y$ with $|x-y|<\delta$ and at the same time $|f(x)-f(y)|\geq\epsilon_0$.
A: If you go for contradiction, it would mean whatever $\delta$  you take, we have  that $\forall$ $\epsilon$ we have $|x-a_n|< \delta \not \Rightarrow |f(x) - f(a_n)| < \epsilon$. Now take $x=b_n$, and since you can take whatever $\delta$ you want, we have that $\lim a_n = \lim b_n$. Now what does that imply?
A: $\impliedby$ fails. On $\mathbb R,$ let $f(x) = x^2.$ Suppose $\lim a_n = \lim b_n.$ Denote the common limit by $L.$ Then by continuity of $f,\lim f(a_n) = f(L) = \lim f(b_n).$ But $f$ is not uniformly continuous on $\mathbb R.$
