I'm trying to understand how an assertion made in the proof of the scalar maximum principle follows from the compactness of the manifold we're working with. The situation is as follows:
I had an idea but I don't know if it pans out: we have that $u_{\varepsilon}(\cdot, 0) > c$ and that for some $(x_0, t_0)$, $u_{\varepsilon}(x_0, t_0) \leq c $. We can then define $t_1$ as the positive infimum over all such times, that is, define:
$$t_1 = \inf_{t \in [0, T)} \{\text{there exists $x \in M$ such that $u_{\varepsilon}(x, t) \leq c$} \}$$
By definition we have that $u_{\varepsilon}(\cdot, t) > c$ for any $t \in [0, t_1)$. Since $M$ is compact, we also know that $u_{\varepsilon}$ takes maximum and minimal values in $M \times [0, t_1]$. I think one can prove that there exists $x_1 \in M$ that satisfies what we want, but I don't know how to guarantee its existence.
Since $u_{\epsilon}>0$ at $t=0$, by the tube lemma and compactness of $M$ it follows that there exists a neighbourhood of type $M \times [0,\delta)$ on which $u_{\epsilon}>c$.
Following a similar idea, take $t_1:=\sup_{t \in [0,T)}\{t \mid \min_{x \in M} u_{\epsilon}(x,t)>c\}$. The above establishes that $t_1 >0$. It is clear that $\min_{x \in M} u_{\epsilon}(x,t_1)\geq c$. If $\min_{x \in M} u_{\epsilon}(x,t_1)$ were greater than $c$, then again by the tube lemma it would follow that $t_1$ is actually not the supremum of the set. Therefore, we have that $\min_{x \in M} u_{\epsilon}(x,t_1)=c$, and by compactness there is some $x_1$ such that $u(x_1,t_1)=c$.