This question regards Advanced Calculus of Several Variables, by C.H. Edwards.
Am I correct that there is an error in Edwards's proof of the following theorem?
Theorem V1-1.2 Let $\mathcal{S}$ be a subset of $\mathbb{R}^{k}$. Let $\left\{ f_{n}\right\} _{1}^{\infty}$ be a sequence of continuous real-valued functions on $\mathcal{S}$. If $\left\{ f_{n}\right\} _{1}^{\infty}$ converges uniformly to the funtion $f:\mathcal{S}\to\mathbb{R},$then $f$ is continuous.
PROOF: Given $\varepsilon>0$, first choose $N$ sufficiently large that $\left|f_{N}\left[x\right]-f\left[x\right]\right|<\frac{\varepsilon}{3}$ for all $x\in\mathcal{S}$ (by the uniform convergence). Then, given $x_{0}\in\mathcal{S}$, choose $\delta>0$ such that
$$ x\in\mathcal{S},\left|x-x_{0}\right|<\delta\implies\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|<\frac{\varepsilon}{3} $$
(by the continuity of $f_{N}$ at $x_{0}$). If $x\in\mathcal{S}$ and $\left|x-x_{0}\right|<\delta$, then it follows that
$$ \left|f\left[x\right]-f\left[x_{0}\right]\right|\le\left|f\left[x\right]-f_{N}\left[x\right]\right|+\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|+\left|f_{N}\left[x_{0}\right]-f\left[x\right]\right|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon, $$
so $f$ is continuous at $x_{0}$.
The attached diagram shows that, if both $\left|f\left[x\right]-f_{N}\left[x\right]\right|>\frac{\varepsilon}{6}$ (magenta) and $\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|>\frac{\varepsilon}{6}$ (blue) , then $\left|f_{N}\left[x_{0}\right]-f\left[x\right]\right|>\frac{\varepsilon}{3}$ (red).
So, it seems to me, Edwards's inequality does not follow from what has been given.
I propose that asserting $\left|f\left[x\right]-f_{N}\left[x\right]\right|<\frac{\varepsilon}{4}$ and $\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|<\frac{\varepsilon}{4}$ does imply $\left|f_{N}\left[x_{0}\right]-f\left[x\right]\right|<\frac{\varepsilon}{2};$ so that
$$ \left|f\left[x\right]-f\left[x_{0}\right]\right|\le\left|f\left[x\right]-f_{N}\left[x\right]\right|+\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|+\left|f_{N}\left[x_{0}\right]-f\left[x\right]\right|<\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{2}=\varepsilon, $$
will repair the proof.
Edit to add a second diagram so the casual reader will not be misled by the first one. Note the color-coding is different than that of the first diagram.
It seems that Edwards made a typo. The last line in the proof should be $$\left|f\left[x\right]-f\left[x_{0}\right]\right|\le\left|f\left[x\right]-f_{N}\left[x\right]\right|+\left|f_{N}\left[x\right]-f_{N}\left[x_{0}\right]\right|+\left|f_{N}\left[x_{0}\right]-f\left[\color{\red}{x_0}\right]\right|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon,$$ because $\left|f_{N}\left[x_{0}\right]-f\left[x_0\right]\right|<\frac{\varepsilon}{3}$ due to uniform convergence (it holds for all $x\in\mathcal{S}$, so in particular for $x_0$.)