Let $X$ be a metric space and let $\{f_n\}$ be a sequence of uniformly continuous functions on $X$. Is it true that if I can show that for any convergent sequence $x_n$, $x_n \rightarrow x$, there exists an N such that for all $n \in \mathbb{N}$ and for all $k > N$, $|f_n(x_k)-f_n(x)| < \epsilon$, then $\{f_n\}$ is equicontinuous?
The reason I think this is true is that, it seems like the choice of $N$ is analogous (or at least very intimately tied to) to the choice of $\delta$ at the limit point. But, most of the proofs that show sequence preservation and epsilon-delta continuity equivalence, don't find the actual delta, so I'm not sure how to prove my idea.
Consider $X = [0,1]$ with the standard metric and $$f_n(x) = \begin{cases} nx\sin\left(\frac{1}{x}\right) & x > 0 \\ 0 & x=0\end{cases}$$
Then all $f_n$ are continuous and because $[0,1]$ is compact in $\Bbb R$ also uniformly continuous but $\{f_n\}$ is not equicontinuous in $0$
Now take $x_n = \frac{1}{n\pi}$ then it holds $f_n(x_k) = 0$ for all $n,k \in \Bbb N$ and so $$|f_n(x_k)-f_n(x)| = 0 < \varepsilon$$ so your condition holds but still: $\{f_n\}$ is not equicontinuous in $0$
The result may differ if you claim that your condition hold for all convergent sequences $(x_n)_{n\in\Bbb N}$ with $x_n \to x$