Let $X$ be a subset of $\Bbb{R}$, and let $f:X\to\Bbb{R}$ be a function. Then the following two statements are logically equivalent:
(a) $f$ is uniformly continuous on $X$.
(b) Whenever $(x_{n})_{n=0}^{\infty}$ and $(y_{n})_{n=0}^{\infty}$ are two equivalent sequences consisting of elements of $X$, the sequences $(f(x_{n}))_{n=0}^{\infty}$ and $(f(y_{n}))_{n=0}^{\infty}$ are also equivalent.
My solution (Edit)
I am mainly interested in the implication $(b)\Rightarrow(a)$.
Let us suppose that $(b)$ holds and $(a)$ does not hold.
The following statements are equivalent
$f:X\to\mathbb{R}$ is uniformly continuous
for every $\varepsilon > 0$, there exists a $\delta > 0$ such that for every $x,y\in X$, \begin{align*} |x - y| < \delta \Rightarrow |f(x) - f(y)| \leq \varepsilon \end{align*}
Based on it, we are going to demonstrate the proposed statement by contradiction.
In other words, let us assume there exists a $\varepsilon > 0$ such that for every $\delta > 0$ there are $x,y\in X$ \begin{align*} (|x - y| < \delta)\wedge (|f(x) - f(y)| > \varepsilon) \end{align*}
In particular, for each $\delta = 1/n$, there are $x_{n},y_{n}\in X$ satisfying \begin{align*} |x_{n} - y_{n}| \leq 1/n\quad\wedge\quad|f(x_{n}) - f(y_{n})| > \varepsilon \end{align*}
Taking the limit, we get that $\displaystyle\lim_{n\rightarrow\infty}(x_{n} - y_{n}) = 0$, but $f(x_{n})$ and $f(y_{n})$ are not equivalent, contradicting the given assumption.
Could someone help me to grasp this difference properly?