Corollary of high-dimensional Mean Value Theorem 
Theorem. Suppose that $G\subset \mathbb{R}^m$ is an open, path-connected set. If the function $f:G\to \mathbb{R}$ is
differentiable in the domain $G$ and its differential equals zero at
every point $x\in G$, then $f$ is constant in the domain $G$,
Proof. We first show that if $x\in G$, then the function $f$ is constant in a ball $B(x;r)\subset G$. Indeed, if $x+h\in B(x;r)$, then
$[x,x+h]\subset B(x;r)\subset G$. Applying high-dimensional version of
Mean Value Theorem, we obtain $$f(x+h)-f(x)=f'(\xi)h=0\cdot h=0,$$
that is, $f(x+h)=f(x)$, and the values of $f$ in the ball $B(x;r)$ are
all equal to the value at the center of the ball.
Now let $x_0,x_1\in G$ be arbitraru points of the domain $G$. By the
path-connectedness of $G$, there exists a path $\Gamma:[0,1]\to G$
such that $\Gamma(0)=x_0$ and $\Gamma(1)=x_1$. Let $B(x_0;r)$ be a
ball with center at $x_0$ contained in $G$. Since $\Gamma(0)=x_0$ and
the mapping $\Gamma(t)$ is continuous, there is a positive number
$\delta_0\in(0,1)$ such that $\Gamma(t)\in B(x_0;r)\subset G$ for
$t\in [0,\delta_0]$. Then, by what has been proved, $(f\circ
 \Gamma)(t)\equiv f(x_0)$ on the interval $[0,\delta_0]$.
Let $\ell=\sup \mathcal{C}$, where
$$\mathcal{C}:=\{\delta\in(0,1]:f(\Gamma(t))=f(x_0) \ \forall t\in
 [0,\delta]\}.$$ Obviously this supremum exists because $\mathcal{C}$
is bounded above and $\mathcal{C}\neq \varnothing$ since $\delta_0\in
 \mathcal{C}$. By the continuity of the function $f(\Gamma(t))$ we have
$f(\Gamma(\ell))=f(x_0)$. But then, $\ell=1$. Indeed, if that were not
so, we could take a ball $B(\Gamma(\ell);r)\subset G$, in which the
value of function $f$ is equal to $f(\Gamma(\ell))=f(x_0)$, and then
by continuity of the mapping $\Gamma$ find $\Delta>0$ such that
$\Gamma(t)\in B(\Gamma(\ell);r)$ for $t\in [\ell,\ell+\Delta]$. But
then $(f\circ \Gamma)(t)=f(\Gamma(\ell))=f(x_0)$ for $t\in
 [0,\ell+\Delta]$, and so $\ell\neq \sup \delta$.
Since $\sup \mathcal{C}=1$, then $f(\Gamma(1))=f(x_0)$ which implies
that $f(x_1)=f(x_0)$. Hence $f\equiv \text{const}$ on $G$.

I think I undestood the technical side of the proof but I need to emphasize it is not so trivial as in one-dimensional setting. However, I am little confused and I am not grasping the general idea behind the proof. Firstly, we have shown that in any small ball inside the $G$ the function is constant and I understood that. However, I have a
Question 1. What is the idea of this $\sup \mathcal{C}$? Can anyone explain it in a simpler way rather than some formula?
Question 2 What is the general idea of this proof? I am confused with this supremum stuff.
In each theorem I am trying to understand the idea and that is why I am asking this question. Thank you so much!
 A: I'll try to explain the idea behind the proof a bit. A picture might also help here (I provide one for $n=2$). So we start by taking two arbitrary points $x_0, x_1 \in G$ and we want to show that $f(x_0)=f(x_1)$. Since $G$ is connected there is a continuous curve $\Gamma: [0,1] \to G$ that connects $x_0$ and $x_1$.
The first part asserts that if we move a little away from $x_0$ along the curve $\Gamma$ the value of $f$ does not change. Then we want to find the most we can move away from $x_0$ along $\Gamma$ such that the value of $f$ remains the same (and equal to $f(x_0)$). By most, I mean in terms of the parameter of $\Gamma$ (and not euclidean distance) which you can think of as time.
This is measured by $\ell$. Now it should be clear that we want $\ell$ to be equal to $1$ for $f(x_0) = f(x_1)$ to hold. If $\ell < 1$ then we can take a ball around $\Gamma(\ell)$ in which $f$ will be constant and equal to $f(\Gamma(\ell)) = f(x_0)$ (this is the same argument as in the first step). But then we could move just a little more along $\Gamma$ without leaving this ball. This implies that  $f(\Gamma(\ell + \Delta))$ would be equal to $f(x_0)$ for some small $\Delta$, contradicting the property of $\ell$. So indeed $\ell = 1$ and $f$ is constant along $\Gamma$ and in particular, $f(x_0)=f(x_1)$.
Here is the drawing, for some geometric intuition:

Remark: There is also a simpler proof provided you are familiar with the fact that an open and connected subset of $\mathbb{R}^n$ is path-connected. In that case, it suffices to prove that $f$ is constant along line segments, which is a direct consequence of the high-dimensional mean value theorem.
