Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$ be bounded and open. We know that if $f\in C^1(\Omega)$ and $\nabla f$ is bounded, then $f$ is Lipschitz continuous (hence uniformly continuous) and hence $f$ has a unique continuous extension to $\overline\Omega$.
Now suppose we only that $f\in C^1(\Omega)$ and $\sup_{x\in K}\|\nabla f(x)\|<\infty$ for all compact $K\subseteq\Omega$. Is this enough to ensure the existence of a unique continuous extension of $f$ to $\overline\Omega$?
As stated, no it's not. For example, $$f : (0, 1) \to \Bbb{R} : x \mapsto \frac{1}{x}.$$ This is continuously differentiable, and the derivative is bounded on each compact subset of $(0, 1)$, but no continuous extension is possible.
(Note that being continuously differentiable implies that the derivative is bounded on each compact subset of the interior of the domain, so functions on open sets were always going to be ripe for picking counterexamples.)