A question on a proof of $C^1$ implies locally Lipschitz I stumbled upon this answer here while studying the proposition that if $f: \mathbb R^n \to \mathbb R^n$ is $C^1$ then $f$ is locally Lipschitz.
The answer in the link applies Taylor's theorem. 
And I was wondering if one could alternatively argue using the Fundamental Theorem of Calculus. Concretely, 

I wrote this proof and was wondering if someone could tell me if it is
  correct?

Proof:
Let $f: \mathbb R^n \to \mathbb R^n$ be $C^1$ and let $x_0 \in \mathbb R^n$. Since $f$ is $C^1$ its derivative $f'$ is continuous. Hence $f'$ attains a maximum $L$ on $\overline{B(x_0,1)}$. Then since
$$ f(x) - f(x_0) \stackrel{FTC}{=} \int_{x_0}^x f'(t) dt$$
we have
$$ \|f(x) - f(x_0)\| = \left \| \int_{x_0}^x f'(t) dt \right \| \le \int_{x_0}^x \|f'(t)\| dt \le L \|x-x_0\|$$
hence $f$ is Lipschitz continuous on $B(x_0, 1)$.
 A: Since $f\in C^1$, there exists a $\delta>0$, such that for $\|\overrightarrow{x}-\overrightarrow{x_0}\|<\delta$, there is
$$
\left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}-f'(\overrightarrow{x_0})\right\|<1
$$
Since 
$$
\left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}-f'(\overrightarrow{x_0})\right\|\geqslant\left\|\dfrac{f(\overrightarrow{x})-f(\overrightarrow{x_0})}{\overrightarrow{x}-\overrightarrow{x_0}}\right\|-\left\|f'(\overrightarrow{x_0})\right\|
$$
There is
$$
\left\|f(\overrightarrow{x})-f(\overrightarrow{x_0})\right\|<(\|f'(\overrightarrow{x_0})\|+1)\|\overrightarrow{x}-\overrightarrow{x_0}\|=L|\overrightarrow{x}-\overrightarrow{x_0}\|
$$
So $f$ is locally Lipschitz.
A: Well, you can choose a convex neigborhood $V$ of $x\in R^n $. Then, for every $y\in V$ you can define $\lambda(t)=f(x+ty)$, with $t \in [0, 1]$. Compute the $\lambda'(t)$, use the chain rule and the mean value inequality for vector valued functions with domain in $R$.
