False proof why $C^1$ implies locally Lipschitz I produced a (possibly) false proof of why $C^1$ implies locally Lipschitz for $f: \mathbb R^n \to \mathbb R^n$.

Please could someone tell me where my mistake is?

Proof:
Let $f: \mathbb R^n \to \mathbb R^n$ be continuously differentiable and let $x_0 \in \mathbb R^n$. The goal is to find a neighbourhood of $x_0$ on which $f$ is Lipschitz continuous. 
Since $f$ is differentiable by the definition of the derivative given $\varepsilon > 0$ there exists $\delta > 0$ such that 
$$ \|x-x_0\| < \delta \text{ implies }\left  \|f'(x_0) - {f(x) - f(x_0) \over x - x_0} \right \| < \varepsilon$$
In particular, there is $\delta$ such that $\left  \|f'(x_0) - {f(x) - f(x_0) \over x - x_0} \right \| < 1$ which implies that for $\|x-x_0\| < \delta$ we have $$ \left  \|  {f(x) - f(x_0) \over x - x_0} \right \| < 1 + \|f'(x_0)\|$$
Hence $B(x_0, \delta)$ is a neighbourhood of $x_0$ on which $f$ is Lipschitz continuous with Lipschitz constant $L=1$. 
 A: I'm pretty sure $C^1$ does imply locally Lipschitz. But the Lipschitz constant is the bound of the (norm of the) derivative, for which you need continuity of the derivative. This follows easily from fundamental theorem of calculus. To see that being $C^1$ does not imply $1$-Lipschitz, consider $f(x)=2x$.
Your argument is flawed from the very beginning: as others have suggested, your notation applies to the case of $n=1$. This can probably be easily fixed. More importantly, you have merely shown that if $\lvert f(x)-f(x_0)\rvert\leq (1+C)\cdot \lvert x-x_0\rvert$, for $C=\lvert f'(x_0)\rvert $, for fixed $x_0$.
I don't think this implies even being locally Lipschitz at $x_0$ (consider something like $f(x)=x\sin(1/x)$), much less being $1$-Lipschitz, and it certainly does not imply being locally Lipschitz everywhere (consider $f(x)=\sin(1/x)$ for $x\neq 0$, $f(0)=0$: this function satisfies your condition anywhere except $0$, but is definitely not Lipschitz, or even continuous, at $0$).
Still, this can easily be extended to give another proof that a $C^1$ function is locally Lipschitz: just let $x_0$ vary, and then notice that the derivative is locally bounded, to see that in a given set the Lipschitz constant is bounded by $1$ plus the bound on derivative. Then let $1$ tend to zero (in the part which says "in particular" in your proof), to obtain the same bound on the Lipschitz constant as the one which you can get from the fundamental theorem of calculus (and which is optimal, just by considering linear functions).
A: Your proof could be correct if you had taken a unit closed ball centered at $x_0$, used mean value inequality, the Lipschitz constant is $L=\sup_{B(x_0,1)}||Df(x)||$.
A: There is no reason to do the quotient $\dfrac{f(x)-f(x_0)}{x-x_0}$ unless your n is 1. Sorry for my English. I think is this.
