@TheBearMinimum, if I am not mistaken you are referring to the proof of Thm. 5.2.3., which was problem 3 on Exam 2 Part 1 (Wed. portion). He gave us a proof in class similar to the following:
Assume that $f$ maps the open interval $I$ to $\mathbb{R}$, that $c \in I$, and that $f$ is differentiable at $c$. Prove that $f$ is continuous at $c$.
proof:
To show $f$ is continuous at $c$, we use Limit Characterization of Continuity and W.T.S. $\lim_{x\rightarrow c} f(x) = f(c)$ or equivalently that $\lim_{x\rightarrow c} (f(x) - f(c)) = 0$ since we can assert by the Algebraic Limit Thm. that:
$$\lim_{x\rightarrow c} (f(x) - f(c)) = \lim_{x \rightarrow c} f(x) - \lim_{x \rightarrow c} f(c) = \lim_{x \rightarrow c} f(x) - f(c)$$
Since $f$ is differentiable, we can use the definition of a derivative and the Algebraic Limit Thm. to assert the following is true:
$\lim_{x\rightarrow c} (f(x) - f(c)) = \lim_{x\rightarrow c} \Bigg(\frac{f(x)-f(c)}{x-c} *(x-c)\Bigg) = \lim_{x\rightarrow c} \Bigg(\frac{f(x)-f(c)}{x-c}\Bigg)*\lim_{x\rightarrow c} (x-c) = f'(c) *0 = 0$
Thus $f$ is continuous at c.