Uniform convergence of $C^1$ functions up to the boundary In Section 1.4 of the book "Partial Differential Equations in Action: From Modelling to Theory" by Sandro Salsa, I encountered the following proposition (rephrased):

Let $\Omega$ be a bounded, open, connected set in $\mathbb{R}^n$.
Consider a sequence of functions $u_m \in C^1(\overline{\Omega})$ (the
notation here means $u_m$ has continuous partials that can be extended
continuously up to $\partial \Omega$). If $u_m(x_0)$ converges at some
$x_0\in \Omega$, and $\nabla u_m$ converges uniformly to some $F \in C^1(\overline{\Omega})$
, then $u_m$ converges uniformly to some $u \in C^1(\Omega)$
, with $\nabla u = F$.

My attempt at proof: The uniform convergence of $\nabla u_m$ implies uniform boundedness of $\nabla u_m$. This would allow proving the equicontinuity of $u_m$ (which is both necessary and sufficient for establishing the result) by computing a line integral, IF the domain $\Omega$ has a nice shape, e.g. convex. For general domain $\Omega$, one could still manage to prove the equicontinuity for some $\epsilon$ distance away from the boundary, by covering with small balls, but it doesn't seem possible to extend it all the way to the boundary. For example, consider the domain in this example.
Question: how does one prove this proposition in general? (Or does it even hold in general?)
 A: A bounded, connected, open subset of $\mathbb{R}^n$ is path-connected.
To prove it, given any point $x \in \Omega$, consider all points that are path-reachable from $x$, and show that the set of all such points, say $R_x$, is nonempty, open, and closed.
Proofs of each part:

*

*Nonempty: $x \in R_x$

*Open: given a path from $x$ to $y$, one can draw an open ball $B_y$ centered at $y$, then for any $z \in B_y$, extend the path from $x$ to $y$ by a line segment from $y$ to $z$ to produce a path from $x$ to $z$.

*Closed: show the complement is open. For any $y \in R_x^c$, draw an open ball $B_y$ centered at $y$. Let $z \in B_y$. Suppose there was a path from $x$ to $z$; then one could extend the path from $x$ to $z$ by a line segment from $z$ to $y$ in order to produce a path from $x$ to $y$, but this is impossible by assumption. So $R_x^c$ is open and thus $R_x$ is closed.

Hence for every $x$, $R_x=\Omega$ which is almost sufficient to prove your claim. (The issue is that for your application, you need some control on the lengths of paths.)
