Assume $\Omega$ is a unit open ball, and $f \in C^1(\Omega) \cap C(\overline{\Omega})$, satisfies $|f(x)|\leq 1$ for all $x$.
How to prove there exists $x_0$ s.t. $|\nabla f(x_0)|\leq 1$?
My idea is to assume all the gradient have norm larger than $1$. Then we create a curve started at origin, and its tangent has direction the same (or opposite direction) as the gradient. Then since the length of the curve must larger than $1$, then we could find a dot $x_1$ s.t. $|f(x_1)|>1$, a contradiction.
However it's hard to describe the curve, and I don't know if it's right.
I think you can formalize your idea the following way: Assume $|\nabla f(x)| > 1$ $(x \in \Omega)$. Now set $$ g:\Omega \to \mathbb{R}^n, \quad g(x)= \frac{\nabla f(x)}{|\nabla f(x)|} $$ and consider the inital value problem $$ u'(t) = g(u(t)), \quad u(0)=0. $$ Since $g$ is continuous on $\Omega$ Peano's Theorem gives a nonextendable solution $u:(t_{\min},t_{\max}) \to \Omega$, and since $|g(x)| = 1$ $(x \in \Omega)$ we have $t_{\min} \le -1$ and $t_{\max} \ge 1$. Moreover, since $|u'(t)| = 1$ $(t \in (t_{\min},t_{\max}))$ the function $u$ is Lipschitz continuous on $(t_{\min},t_{\max})$ and therefore can be continuously extended to $u:[t_{\min},t_{\max}] \to \overline{\Omega}$. Now $$ \int_{t_{\min}+\varepsilon}^{t_{\max}-\varepsilon} |\nabla f(u(t))| dt = \int_{t_{\min}+\varepsilon}^{t_{\max}-\varepsilon} \langle \nabla f(u(t)),u'(t) \rangle dt = f(u(t_{\max}-\varepsilon))-f(u(t_{\min}+\varepsilon)), $$ and $\varepsilon \to 0+$ yields $$ 2<\int_{t_{\min}}^{t_{\max}} |\nabla f(u(t))| dt = f(u(t_{\max}))-f(u(t_{\min})). $$ Thus $|f(u(t_{\max}))| > 1$ or $|f(u(t_{\min}))| > 1$.