Is it true that if $f:O_{\text{open}}\subset\Bbb R\rightarrow\Bbb R$ is continuous then it is locally Lipschitz?
Maybe the function $f(x)=\sin(1/x)\cdot x$ is a counter example because at $x=0$ we can find a sequence $x_n\rightarrow 0$ such that $|f'(x_n)|\rightarrow\infty$ so there is no neighborhood of $0$ such that Lipschitz condition holds.
Is there an additional condition we could impose to make the statement true?