Baby Rudin Chapter 4 Exercise 1

The mistake is in the line

Let $$\lim\limits_{h\rightarrow 0}{f(x+h)} = \lim\limits_{h\rightarrow 0}{f(x-h)} = v = f(x).$$

If $\lim\limits_{h\rightarrow 0}{f(x+h)}$ did exist, then it is true that it is equal to $\lim\limits_{h\rightarrow 0}{f(x-h)}$. However, there is no reason it should exist, and even if it did exist, there is no reason it has to equal $f(x)$.

An example of where the limit does not even exist is the function $$ f(x) = \left\{\begin{matrix} \frac{1}{x^2}& x\ne 0 \\ 0 & x = 0\end{matrix}\right.$$

This is false. Take for example $f(x)=0$ for $x\neq0$ and $f(0)=1$.

The problem is in the second equation of your proof. The limit $v$ needn't be $f(x)$. In fact, $f$ is continuous if and only if the limit $v$ exists and $v=f(x)$.