Equivalence of uniform continuity and a relaxation of Cauchy continuity Let $f:(a,b)\to\mathbb{R}$, then $f$ is uniform continuous iff it is Cauchy continuous, i.e. for every pair of sequences $\{x_n\}$ and $\{y_n\}$ such that $\lim_{n\to \infty }|x_{n}-y_{n}|=0$, we have $\lim_{n\to \infty }|f(x_{n})-f(y_{n})|=0$.
What if we change $\lim_{n\to \infty }|f(x_{n})-f(y_{n})|=0$ in the Cauchy continuity to
$$\lim_{n\to \infty }|f(x_{n})-f(y_{n})|=0\text{ or}\lim_{n\to \infty }|f(x_{n})-f(y_{n})|\text{ does not exist}?$$
The "only if" part still holds because we have in fact weakened the Cauchy continuity, but what about the "if" part?
To be specific, I'll restate the question:

For an $f:(a,b)\to\mathbb{R}$ consider the following condition:
(*) for every pair of sequences $\{x_n\}$ and $\{y_n\}$, if $\lim_{n\to \infty }|x_{n}-y_{n}|=0$ and $\lim_{n\to \infty }|f(x_{n})-f(y_{n})|=M$ for some $M\in\mathbb{R}$, then $M=0$.
Prove/Disprove: If $f$ satisfies (*) then it is uniformly continuous?

 A: The answer to your question is affirmative, but the proof turned out to be non-trivial. For instance, I used Baire and Ramsey theorems. Let’s begin. 
For each natural $n$ put $X_n=\{x\in (a,b):|f(x)|\le n\}$. We claim that the set $X_n$ is a closed subset of $(a,b)$. Indeed, suppose to the contrary that there exists a point $x\in ((a,b)\cap\overline{X_n})\setminus X_n$. There exists a sequence $\{x_i\}$ of points of $X_n$, converging to $x$. Since a sequence $\{f(x_i)\}$ is bounded, there exists a subsequence $\{x_{i(k)}\}$ of $\{x_i\}$ such that a sequence $\{f(x_{i(k)})\}$ converges to a number $t$. Clearly, $|t|\le n$ and if $t\ne f(x)$ then a pair $\{x_{i(k)}\}$ and $\{x’_k=x\}$ of sequences violates the property, a contradiction.  
For each natural $n$ put $X’_n=\operatorname{int} X_n$. By zhw.’s answer, $f$ is continuous on each $X’_n$. Put $X’=\bigcup_n X’_n$. Then $X’$ is a union of a family $\mathcal U$ of its maximal open intervals. Let $U\in\mathcal U$ be any interval and $x\in U$ be any point. Since $U=\bigcup_n U\cap X’_n$, there exists $n$ such that $x\in U\cap X’_n$. It follows that $f$ is continuous at $x$, so $f$ is continuous on $U$ (and on $X’$).  
We claim that $f$ is bounded on $U$. Indeed, suppose to the contrary that there exists a sequence $\{x_i\}$ of points of $U$ such that $f(x_{i+1})\ge f(x_i)+1$ for each $i$ (or resp. $f(x_{i+1})\le f(x_i)-1$ for each $i$). By this answer,  there exists a monotone subsequence $\{x_{i(k)}\}$ of $\{x_i\}$. For each natural number $l$ let $k$ is the largest number such that $f(x_{i(k)})\le f(x_1)+l$  (resp. $f(x_{i(k)})\ge f(x_1)-l$) and $k’$ be the smallest number such that $f(x_{i(k’)})\ge f(x_1)+l$  (resp. $f(x_{i(k’)})\le f(x_1)-l$). Since the function $f$ is continuous on $U$, by the intermediate value theorem there exists a point $y_l$ in a closed interval between $x_{i(k)}$ and $x_{i(k’)}$, such that $f(y_l)=f(x_1)+l$ (resp. $f(y_l)=f(x_1)-l$). Then $\{y_l\}$ is a Cauchy sequence, because $\lim_{l\to\infty} y_l=\lim_{k\to\infty} x_{i(k)}$ and a pair $\{y_k\}$ and  $\{z_k=y_{k+1}\}$ of sequences violates the property, a contradiction.
Put $Y=(a,b)\setminus X’$. If $Y$ is empty then $\mathcal U=\{(a,b)\}$, so $f$ is bounded and, by zhw.’s answer, uniformly continuous. 
Suppose to the contrary that $Y$ is non-empty. The topological space $Y$ is a union $\bigcup_n Y\cap X_n$ of a family of its closed subsets. Since $Y$ is a closed subset of a topological space metrizable by a complete metric, it satisfies the condition of Baire theorem. So there exists a number $m$ such that a set $Y\cap X_m$ has non-empty interior in $Y$.  That is there exists an interval $(c,d)$ such that $(c,d)\cap Y$ is a non-empty subset of $X_m$. Put $\mathcal U’=\{U\cap (c,d):U\in\mathcal U\}$.
We claim that $(c,d)\subset X_k$ for some natural $k$. Indeed, otherwise we inductively construct an increasing sequence $\{k_n\}$ of natural numbers and a pair $\{x_n\}$ and $\{y_n\}$ of sequences violating the property as follows. Put $k_1=m$. For any $n$, given $k_n$, pick an arbitrary point $z_n\in (c,d)\setminus X_{k_n}$. Since $z_n\not\in X_m$, $z_n\in U’_n$ for some $U’_n\in\mathcal U’$. Then $U’_n=U_n\cap (c,d)$ for some $U_n\in\mathcal U$. By the above, there exists a number $k_{n+1}>k_n$, such that $U_n\subset X_{k_{n+1}}$. Since $(c,d)\not\subset U_n$, the interval $U_n$ (and so the interval $U’_n$ too) has an endpoint $x_n\in (c,d)$. Then $x_n\in Y\cap (c,d)\subset X_m$. Since the function $f$ is continuous on $U’_n\cup \{x_n\}$, by the intermediate value theorem there exists a point $y_n$  belonging to a segment between $x_n$ and $z_n$ such that $|f(x_n)-f(y_n)|=1$. Since $x_n$ is an endpoint of $U’_n$  and $y_n\in U’_n$, we have that $|x_n-y_n|$ is not bigger than the length $|U’_n|$ of the interval $U’_n$. Since for each $n\ge 1$ we have 
$z_n\in U_n\setminus X_{k_n}$, whereas $U_p\subset X_{k_{p+1}}\subset X_{k_n}$ for each $1\le p<n$, the intervals $U_n$ are distinct and so, by their maximality, mutually disjoint. Then  we have $\sum |U’_n|\le \sum |U_n|\le |(a,b)|=b-a$, so both  sequences $\{|U’_n|\}$ and $\{|x_n-y_n|\}$ converge to zero. $\square$
A: Here's a proof for bounded $f$ having the given property: Suppose $f$ is not uniformly continuous. Then there exists $\epsilon>0$ such that no $\delta >0$ "works". Thus for $\delta = 1/n, n=1,2,\dots,$ there exist $x_n,y_n\in (a,b)$ such that $|x_n-y_n|<1/n$ and $|f(x_n)-f(y_n)|\ge \epsilon.$
Because $f$ is bounded, there exists a subsequence $n_k$ of $1,2,\dots$ such that both $f(x_{n_k}),f(y_{n_k})$ converge. We then have $|x_{n_k}-y_{n_k}|\to 0$ and $|f(x_{n_k})-f(y_{n_k})|\to L$ for some $L\ge \epsilon.$ That contradicts the assumption on $f.$
Thus $f$ is uniformly continuous as desired.
A: It is true that $f$ is uniformly continuous on $(a,b)$ -- finite interval -- if and only if we can assign (finite) values to $f(a)$ and $f(b)$ while retaining continuity. And this is equivalent to the existence and finiteness of limits at the endpoints.
The following is a contrapositive proof of the last statement in the post.
So if $f$ is not uniformly continuous then 
a) either $\limsup_{x \to a} f(x)$ is $+ \infty$ or $\liminf_{x\to a} f(x) = - \infty $; or
b) $t:=\liminf_{x \to a} f(x) < \limsup_{x \to a} f(x) =:s $ with both finite.
In either case you can find subsequences $x_n$ and $y_n$ that both converge to $a$, and hence $|x_n - y_n|$ converges to zero, but $|f(x_n) - f(y_n)|$ converges to $M$ with $M \neq 0$. Let me show how:
In case b) there exists, by definitions, sequences $x_n \to a $ and $y_n \to a$ such that $f(x_n) \to s$ and $f(y_n) \to t$, thus $|f(x_n)-f(y_n)| \to s-t \neq 0$.
In case a) I will only do $\limsup = +\infty$ case as the other is similar. We can easily see that usual continuity of $f$ over $(a,b)$ easily follows from the |weak Cauchy" condition( Mini proof: take $y_n$ to be the constant sequence $y_n =x $ and $x_n$ any other seq converging to $x$.) Now, take a seq $x_n$ that decreases to $a$ and such that $\lim f(x_n) \to +\infty$. We can further require that $f(x_{n+1}) > f(x_n) + 1$, i.e gaps of at least 1 unit. By intermediate value theorem, for each $n$ we can find $y_n \in (x_{n+1},x_n)$ such that $f(y_n) = f(x_n) + 1$ -- precisely. Now, of course $y_n$ also converge to $a$, hence $|y_n - x_n| \to 0$, but their $|f(y_n) - f(x_n)| =1 $ for all $n$!
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