Let $I$ be an interval, $c \in I$, and $f: I \rightarrow \mathbb{R}$.
Consider the following definitions:
(1) For every sequence $\left(x_{n}\right)_{n \in \mathbb{N}}$ contained in I such that $\left(x_{n}\right)_{n \in \mathbb{N}}$ converges to c, the sequence $\left(f\left(x_{n}\right)\right)_{n \in \mathbb{N}}$ converges to $f(c)$
(2) $f$ is continuous at $c$ (in the delta-epsilon sense)
I am trying to show that $(1)\implies (2)$
Construct an $\{x_{n}\}_{n\in\mathbb{N}}$ such that $x_{n}\in B_{1/n}(c)$, for each $n$. Clearly, $\{x_{n}\}\rightarrow c$, and so $f(x_{n})\rightarrow c$. Thus, there exists $N\in \mathbb{N}$ such that $n>N \implies$ $|f(x_n)-f(c)|<\epsilon$. So, set $\delta :=\frac{1}{N+1}$. Then, $|x-c|<\delta$ $\implies$ $x \in B_{\frac{1}{N+1}}(c)\subset B_{\frac{1}{N}}(c)$, so $|f(x)-f(c)|<\epsilon.$
Proof by contradiction. We shall prove prove that: $$ \text{Not}\,(2) \quad\Longrightarrow\quad \text{Not}\,(1). $$ If not (2), then there exists an $\varepsilon>0$, such that, for all $\delta>0$, there exists a $y_\delta\in I$, with $$ |y_\delta-c|<\delta \quad\text{and}\quad |f(y_\delta)-f(c)|\ge\varepsilon. $$ In particular, for $\delta=1/n$, define $x_n=y_{1/n}$.
Clealy, $|x_n-c|<1/n$, while $|f(x_n)-f(c)|\ge \varepsilon$, which contradicts (1).