Smooth function on bounded domain is Lipschitz

I am unsure, if my proof is correct. Would anybody please verify it?

Let $f:\mathbb R^n \to \mathbb R$ be smooth, and let $U\subset R^d$ be a bounded domain, i.e., $U$ is open, connected and bounded.

Hence the closure $cl(U)$ is closed, bounded and connected. As the restriction$f|_{cl(U)}$ is smooth on $cl(U)$, it is Lipschitz.

Moreover $U\subset \overline U$, $f|_{U}$ is Lipschitz as well.