Let $f:N\to R^n$ be a locally Lipschitz function, where $N$ is an open subset of $R^n$. Let $K$ be a compact set in $N$. Prove that the restriction $f|_K$ of $f$ to $K$ is globally Lipschitz.
I just need to know how I can prove that if $f$ is locally Lipschitz, then $f$ is continuous on $N$ and hence it is also continuous on $K$.
Let $\epsilon>0$ and $x\in N$. Since $f$ is locally Lipschitz, we have some open ball $B$ of radius $\delta$ around $x$ such that $f$ is Lipschitz on $B$. Thus we have some $C$ such that $y\in B\implies |f(x)-f(y)|\leq C|x-y|$. By shrinking $\delta$, we can assume $\delta<\epsilon/C$, thus $$|x-y|<\delta\implies |f(x)-f(y)|\leq C|x-y|< C\delta<\epsilon$$ thus $f$ is continuous on $N$.