Let $f:\mathbb{R}^n\to\mathbb{R}^n$, let $A\subseteq\mathbb{R}^n$ be compact and suppose $\nabla f=0$ on $A$. Show that there is a constant $B$ such that $|f(x)-f(z)|\leq B|x-z|^2$ for all $x,z\in A$.
My attempt: Cover $A$ by finitely many open balls $B_{\delta}(x_1),...B_{\delta}(x_q)$ of radius $\delta$. Since open balls are convex, if $x,z\in B_{\delta}(x_j)\cap A$ for some $j$, then $|f(x)-f(z)|=0$ by the mean value theorem. There is an $\epsilon$ such that if $x,z$ are not in the same ball, then $|x-z|\geq\epsilon$. Then $|f(x)-f(z)|\leq||f(A)||=||f(A)||\frac{1}{\epsilon^2}|x-z|^2$, where $|f(A)|$ is the diameter of $f(A)$. So setting $B=|f(A)|\frac{1}{\epsilon}^2$ gives the result.
Does this proof look correct?
No, I don't agree with your proof. You can cover $A$ by balls as you say, but in general $A$ will be properly contained in the balls. So although the ball $B_\delta(x_j)$ is convex, the set $B_\delta(x_j) \cap A$ need not be, nor even connected. Hence you cannot conclude that $f$ is constant on $B_\delta(x_j) \cap A$.
In fact, I think this claim is false. Take $n=1$ and start with the function $f(x) = |x|^{3/2}$, which is differentiable at $x=0$ with $f'(0)=0$. Consider the sequence $x_n = 1/n$ and choose neighborhoods $U_n$ of the $x_n$ which are disjoint. Now modify $f$ inside each $U_n$ so that it is constant on a smaller neighborhood of $x_n$, without changing the value of $f(x_n) = |x_n|^{3/2}$. We can do this in such a way that $|f(x)| \le 2|x|^{3/2}$ everywhere, and this will imply that $f$ is still differentiable at $x=0$ with $f'(0)=0$. Now $f'$ vanishes on the set $A = \{x_1, x_2, x_3, \dots, 0\}$ which is compact, but $f(x) = |x|^{3/2}$ for every $x \in A$, and this is not 2-Holder continuous.