[proofplan] We fix a point $a \in U$ and define the level set $S = \{x \in U : f(x) = f(a)\}$. We show $S$ is simultaneously open and closed in $U$. Openness follows from the [Mean Value Inequality](/theorems/328) applied on small balls (where $M = 0$ gives $f(y) = f(x_0)$). Closedness follows from [continuity](/page/Continuity) of $f$ (a consequence of [differentiability](/page/Derivative)). Since $U$ is connected and $S$ is nonempty, $S = U$. [/proofplan] [step:Define the level set and show it is open via the Mean Value Inequality] Fix $a \in U$ and define $S = \{x \in U : f(x) = f(a)\}$. We have $a \in S$, so $S \neq \varnothing$. [claim:$S$ is open in $U$] If $x_0 \in S$, there exists $\delta > 0$ with $B_\delta(x_0) \subseteq S$. [/claim] [proof] Since $U$ is open, choose $\delta > 0$ with $B_\delta(x_0) \subseteq U$. For any $y \in B_\delta(x_0)$, the line segment $[x_0, y]$ lies in $B_\delta(x_0) \subseteq U$ (open balls in $\mathbb{R}^m$ are convex). Since $Df_z = 0$ for all $z \in U$, the hypothesis of the [Mean Value Inequality](/theorems/328) holds with $M = 0$ on $[x_0, y]$: \begin{align*} |f(y) - f(x_0)| \leq 0 \cdot |y - x_0| = 0. \end{align*} Therefore $f(y) = f(x_0) = f(a)$, so $y \in S$. [/proof] [/step] [step:Show $S$ is closed in $U$ via continuity of $f$] [claim:$S$ is closed in $U$] If $(x_k)$ is a [sequence](/page/Sequence) in $S$ with $x_k \to x_\infty \in U$, then $x_\infty \in S$. [/claim] [proof] Since $f$ is [differentiable](/page/Derivative) at $x_\infty$, it is [continuous](/page/Continuity) there by [Differentiability Implies Continuity](/theorems/322). Therefore $f(x_\infty) = \lim_{k \to \infty} f(x_k) = f(a)$, so $x_\infty \in S$. [/proof] [/step] [step:Conclude $S = U$ from connectedness] The set $S$ is nonempty (since $a \in S$), open in $U$, and closed in $U$. Since $U$ is connected, the only subsets of $U$ that are both open and closed in $U$ are $\varnothing$ and $U$ itself. Therefore $S = U$, meaning $f(x) = f(a)$ for all $x \in U$. [/step]