Proof of Differentiability at an Interior Point Implies Continuity There

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, so that $d_{\mathbb{R}}(s,t)=|s-t|$ for all real $s,t$ by The Absolute Value Metric on the Real Line.

Write $L=f'(x_0)$. Applying the definition of the derivative with $\varepsilon=1$, there is a real $\delta_1>0$ such that every real $h$ with $0<|h|<\delta_1$ and $x_0+h\in I$ satisfies

For such $h$, the triangle inequality for the absolute value gives

and multiplying by $|h|>0$, together with $|f(x_0+h)-f(x_0)|=|h|\cdot\bigl|\frac{f(x_0+h)-f(x_0)}{h}\bigr|$,

Now let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$ be given. Since $|L|\ge0$, the number $|L|+1$ is positive, so

is a positive real. Let $y\in I$ satisfy $d_{\mathbb{R}}(x_0,y)<\delta$, that is $|y-x_0|<\delta$, and put $h=y-x_0$, so that $x_0+h=y\in I$.

If $h=0$ then $y=x_0$ and $d_{\mathbb{R}}(f(y),f(x_0))=0<\varepsilon$. If $h\ne0$ then $0<|h|<\delta\le\delta_1$, so $(*)$ applies and

In both cases $d_{\mathbb{R}}(f(y),f(x_0))<\varepsilon$. Since $\varepsilon>0$ was arbitrary, $f$ is continuous at $x_0$ relative to $I$. $\blacksquare$