Differentiability implies continuity - A question about the proof

Technically, there is an implicit issue of existence of limits which is being swept under the rug in the presentation you have given. The assumption of differentiability at $x_0$ says that the limit

$$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$

exists as a finite number. The limit $\lim_{x \to x_0} x-x_0$ exists and is zero regardless of our assumptions. Then the product rule for limits tells us both that $\lim_{x \to x_0} f(x)-f(x_0)$ exists, and that it is the product of the two limits above, which means it must be zero. Because the product rule also tells us that the limit exists, we do not have to assume continuity first.