Proof of Differentiability at a Point Implies Continuity There

By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval and $c$ is an interior point of it, so differentiability at $c$ is meaningful. Write $L=g'(c)$, let $|\cdot|$ be the absolute value, and recall from The Absolute Value Metric on the Real Line that $d_{\mathbb{R}}(s,t)=|s-t|$. Claim numbers below refer to Elementary Order Arithmetic in an Ordered Field and to Properties of the Absolute Value in an Ordered Field as indicated.

Set $C=|L|+1$. Since $0\le|L|$ by claim 1 of Properties of the Absolute Value in an Ordered Field and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 3 of the latter gives $0<C$. Hence $C^{-1}$ exists and $0<C^{-1}$ by claim 7.

Step 1 (a Lipschitz bound near $c$). Apply differentiability of $g$ at $c$ with the value $1$ in place of $\varepsilon$: there is $\delta_{1}$ with $0<\delta_{1}$ such that every $k\in\mathbb{R}$ with $0<|k|<\delta_{1}$ and $c+k\in(p,q)$ satisfies

Fix such a $k$ and write $Q$ for the quotient. By claim 5 of Properties of the Absolute Value in an Ordered Field,

and $|Q-L|+|L|<1+|L|=C$ by claim 1 of Elementary Order Arithmetic in an Ordered Field; so $|Q|<C$ by claim 2. Since $k\ne0$ we have $0<|k|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so claim 10 gives $|Q|\,|k|<C\,|k|$. By claim 4 of Properties of the Absolute Value in an Ordered Field and the identity $Q\,k=g(c+k)-g(c)$,

Step 2 (continuity). Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Claim 5 gives $0<C^{-1}\varepsilon$. By claim 9 there is $\delta\in\mathbb{R}$ with $\delta\le\delta_{1}$, $\delta\le C^{-1}\varepsilon$, and $\delta$ equal to $\delta_{1}$ or to $C^{-1}\varepsilon$; in either case $0<\delta$.

Let $y\in(p,q)$ satisfy $d_{\mathbb{R}}(c,y)<\delta$, that is $|y-c|<\delta$. If $y=c$, then $|g(y)-g(c)|=|0|=0<\varepsilon$ by claim 1 of Properties of the Absolute Value in an Ordered Field. Otherwise set $k=y-c$; then $k\ne0$, $c+k=y\in(p,q)$, and $0<|k|<\delta$, so $|k|<\delta_{1}$ by claim 2. Step 1 gives $|g(y)-g(c)|<C\,|k|$. Also $|k|<C^{-1}\varepsilon$ by claim 2, so claim 10 with multiplier $C$ gives $C\,|k|<C\,(C^{-1}\varepsilon)=\varepsilon$, and claim 2 gives $|g(y)-g(c)|<\varepsilon$, that is $d_{\mathbb{R}}(g(y),g(c))<\varepsilon$.

This is exactly the defining condition of continuity of $g$ at $c$ relative to $(p,q)$ as a map into $(\mathbb{R},d_{\mathbb{R}})$.