Function of class $C^1$ implies locally Lipschitz Let $A$ be open in $\mathbf{R}^{m} ;$ let $g: A \rightarrow \mathbf{R}^{n} .$ If $S$ is a subset of $A,$ we say that $g$ satisfies the Lipschitz condition on $S$ if the function
$$
\lambda(\mathbf{x}, \mathbf{y})=|g(\mathbf{x})-g(\mathbf{y})| /|\mathbf{x}-\mathbf{y}|
$$
is bounded for $\mathbf{x}, \mathbf{y}$ in $S$ and $\mathbf{x} \neq \mathbf{y}$. We say that $g$ is locally Lipschitz if each point of $A$ has a neighborhood on which $g$ satisfies the Lipschitz condition.
How do I go about showing that if $g$ is of class $C^{1}$, then $g$ is locally Lipschitz?
I know that the derivatives of $g$ are locally bounded and then I can apply the MVT, but am not sure how to formulate this. The $1-D$ case is easy, but I am struggling with the multidimensional case given.

There is Taylor's theorem for multivariable functions, which gives a mean value estimate. But in this case, it is easy to derive the estimate from scratch. Let $M > 0$ and suppose that for all $x$ in some open convex set $U$, $||Dg(x)|| \leq M$. Let $x_1, x_2 \in U$. Define $\phi \colon [0, 1] \to \mathbb{R}^n$ by $\phi(t) = g(x_1 + t(x_2 - x_1))$. By the chain rule, $\phi'(t) = Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)$. By the fundamental theorem of calculus (applied componentwise),
\begin{align} 
||g(x_2) - g(x_1)|| &= ||\phi(1) - \phi(0)|| \\
&= ||\int_{0}^{1}Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)\,dt|| \\
&\leq \int_{0}^{1}||Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)||\,dt \\
&\leq \int_{0}^{1}||Dg(x_1 + t(x_2 - x_1))|| \cdot ||x_2 - x_1||\,dt \\
&\leq M||x_2 - x_1||.
\end{align}
