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This question is related to the previous one.

Let $\Omega$ be an open connected subset of ${\bf R}^n$ and $f:\Omega\to{\bf R}^n$ be continuously differentiable such that $$ \sup_{x\in\Omega}||Df(x)||<+\infty. $$ Show that for every compact $K\subset \Omega$, there exists $L_K$ such that for every $x,y\in K$, $$ |f(x)-f(y)|\leq L_K|x-y|. $$

Here are some thoughts about it. If $\Omega$ is convex, then one can apply the theorem quoted in the previous question. Suppose it is not. By the compactness of $K$, one can cover it with finitely many open balls contained in $\Omega$. Since open balls are convex, one can have a Lipschitz constant $L_i$ for each ball. For any $x,y\in K$, if they are in the same open ball, then one can bound $|f(x)-f(y)|$ using the maximum of $\{L_i\}$. But when they are not in the same ball, I don't see how to go on.

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The assumption may be different but the same proof works.

Suppose $f$ is not Lipschitz on $K$. Then there exist two sequences $x_n,y_n\in K$ such that $$\lim_{n\to\infty}\frac{|f(x_n)-f(y_n)|}{|x_n-y_n|}= \infty \tag{1}$$ Passing to a subsequence, we can assume $x_n\to x\in K$.

Since $f$ is bounded on $K$, the numerator in (1) is bounded; therefore the denominator tends to zero. This means $y_n\to x$. So, for sufficiently large $n$ the points $x_n,y_n$ both lie in an open ball $B$ centered at $x$ and such that $\overline{B}\subset \Omega$. But $B$ is convex, which allows for the argument you mentioned: $$|f(x_n)-f(y_n)|\le |x_n-y_n|\sup_{\overline{B}} \|Df\|$$ a contradiction.

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