Suppose $f$ is a differentiable function from $\mathbb{R}^n \rightarrow\mathbb{R}^n$. In the case that $n=1$ we know, by the mean value theorem, that if $\sup_x |f'(x)| \le K$, then $f$ is a Lipschitz function with constant $K$. I am intersted in a result like this for $n>1$.
In that case $Df(x)$ is a $n\times n$ matrix. Is there a matrix norm $\|\cdot\|$ that gives the following result: $$ \sup_{x\in\mathbb{R}^n} \|Df(x)\| \le K \quad\Rightarrow\quad \sup_{x,y\in\mathbb{R}^n} \frac{\|f(x) - f(y)\|_2}{\|x-y\|_2} \le K, $$ where $\|\cdot\|_2$ denote the Euclidean norm on $\mathbb{R}^n$. Specifically, is this result true for the spectral radius, that is, the operator norm induced by $\|\cdot\|_2$.
Yes it is.
Consider a differentiable function $f \colon \mathbb R^n \to \mathbb R^m$. As you observed, for $n = 1$ one can use the mean value theorem for each component of $f$ to obtain $$ \left\| \frac{f(x)-f(y)}{x-y} \right\|_2 = \| f'(\theta) \|_2 \le K. $$
Now consider arbitrary $n$. It is enough to show that $f$ is $K$-Lipschitz on any line $\{ x_0+tv : t \in \mathbb R \}$, $\|v\|_2 = 1$. In other words, that $$ g_{x_0,v} \colon \mathbb R \to \mathbb R^n, \quad g_{x_0,v}(t) = f(x_0 + tv) $$ is $K$-Lipschitz. Since $g_{x_0,v}$ is differentiable and $g_{x_0,v}'(t) = Df(x_0+tv) \cdot v$, we have $$ \| g_{x_0,v}'(t) \|_2 \le \| Df(x_0+tv) \|_{\mathbb R^n \to \mathbb R^m} \cdot \| v \|_2 \le K, $$ thus $g_{x_0,v}$ is $K$-Lipschitz by the previous reasoning.