Proof of Continuity at a Point of an Interval in Terms of the Limit

Throughout, continuity at $c$ is that of The Real Line: Standing Notation and Background for Calculus on Intervals (v) and the limit is that of Limit of a Real Function at a Point of an Interval, both taken relative to the interval $I$.

Continuity implies the limit statement. Suppose $f$ is continuous at $c$, and let $\varepsilon>0$. By continuity there is $\delta>0$ such that every $x\in I$ with $|x-c|<\delta$ satisfies $|f(x)-f(c)|<\varepsilon$. If $x\in I$ satisfies $0<|x-c|<\delta$, then in particular $|x-c|<\delta$, so $|f(x)-f(c)|<\varepsilon$. Thus the real number $f(c)$ has the property required in Limit of a Real Function at a Point of an Interval (i); by Uniqueness of the Limit of a Real Function at a Point of an Interval (ii) it is the only real number with that property, and therefore $\lim_{x\to c}f(x)=f(c)$.

The limit statement implies continuity. Suppose $\lim_{x\to c}f(x)=f(c)$, and let $\varepsilon>0$. Choose $\delta>0$ such that every $x\in I$ with $0<|x-c|<\delta$ satisfies $|f(x)-f(c)|<\varepsilon$. Now let $x\in I$ satisfy $|x-c|<\delta$. If $x=c$, then $|f(x)-f(c)|=|0|=0<\varepsilon$. If $x\ne c$, then $|x-c|>0$, so $0<|x-c|<\delta$ and hence $|f(x)-f(c)|<\varepsilon$. In both cases $|f(x)-f(c)|<\varepsilon$, so $f$ is continuous at $c$.