Let $U \subseteq \mathbb{R}^{n}$ be an open and connected set. If $f: U \rightarrow \mathbb{R}$ has partial derivatives equal to zero on every point of $U$, then $f$ is constant.
My idea to prove the result above is to fix $a := (a_1, \ldots, a_n)$, take an arbitrary point $b := (b_1, \ldots, b_n)$, create a chain of points that connect $a$ to $b$ and apply the following version of the Mean Value Theorem:
(Theorem) Let $U \subseteq \mathbb{R}^{n}$ be an open set and $f: U \rightarrow \mathbb{R}$. Suppose that $[a, a+v] \subseteq U$, where $v \in \mathbb{R}^{n}$, and $\partial_{v}f(x)$ exists for every $x \in (a, a+v)$. Then, there exists $\theta \in (0,1)$ such that $f(a+v) - f(a) = \partial_{v}f(a+ \theta v)$.
This seems to be the correct approach. However, for that to work, the chain of points that we take to connect $a$ and $b$ have to move only in the directions $e_1, \ldots, e_n$, one at a time. I know that a connected set in $\mathbb{R}^{n}$ is polygonally connected, but that doesn't say we can connect two points only through vectors of the form $v := te_i$. It makes sense intuitively, and it probably works, but I was unable to find any result about this. Is that correct?