Let $X$ be a subset of $\mathbb{R}$, and let $f : X\to \mathbb{R}$ be a function. Then the following two statements are logically equivalent:
(a) $f$ is uniformly continuous on $X$.
(b) Whenever $(x_n)$ and $(y_n)$ are two equivalent sequences consisting of elements of $X$, the sequences $(f(x_n))$ and $(f(y_n))$ are also equivalent.
Proof
First I will state the definitions of uniform continuity and equivalent sequences.
(Uniform continuity). Let $X$ be a subset of $\mathbb{R}$, and let $f : X\to\mathbb{R}$ be a function. We say that $f$ is uniformly continuous if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $f(x)$ and $f(x_0)$ are $\epsilon$-close whenever $x, x_0 \in X$ are to points in $X$ which are $\delta$-close.
.
(Equivalent sequences). Let $m$ be an integer, let $( a_n)_{n=m}^\infty$ and $( b_n)_{n=m}^\infty$ be two sequences of real numbers, and let $\epsilon> 0$ be given. We say that $( a_n)_{n=m}^\infty$ is $\epsilon$-close to $( b_n)_{n=m}^\infty$ iff $a_n$ is $\epsilon$-close to $b_n$ for each $n\geq m$. We say that $( a_n)$ is eventually $\epsilon$-close to $( b_n)$ iff there exists an $N\geq m$ such that the sequences $(a_n)$ and $(b_n)$ are $\epsilon$-close. Two sequences $(a_n)$ and $(b_n)$ are equivalent iff for each $\epsilon> 0$, the sequences $(a_n)$ and $(b_n)$ are eventually $\epsilon$-close.
since $x\in X$ is an adherent point to $X$, then there exists a sequence $(a_n)$ such that $a_n \in X$ and converges to x. since f is continuous, then the sequence $(f(a_n))$ converges to $f(x)$.
Let $(b_n)$ be a sequence equivalent to $(a_n)$. Therefore, $\forall \epsilon>0, \exists N\text{ such that } |a_n-b_n|\leq\epsilon$. choose $\epsilon=\delta$, then we have $|a_n-b_n|\leq\delta \forall n\geq N$
Hence, $|f(a_n)-f(b_n)|\leq\epsilon\text{ }, \forall n\geq N$
hence, $(f(a_n))$ and $(f(b_n))$ are equivalent.
Is my proof correct?