Why -since this set is also closed in $E$, and $E$ is connected, it follows that it must be all of $E$-?

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If $\bf{f}$ is a differentiable mapping of a connected open set $E \subset \mathbb{R}^n$ into $\mathbb{R}^m$, and if $\mathbf{f}'(\mathbf{x}) = \mathbf{0}$, show that $\mathbf{f}(\mathbf{x})$ is constant in $E$.

The answer is given here. The question that I have about the answer provided is:

Why can we make the following statement (last sentence)?

Since this set is also closed in $E$, and $E$ is connected, it follows that it must be all of $E$.

Can someone provide an proof for this statement based on the definition of what it means for $E$ to be connected, i.e. not a union of two nonempty separated sets.

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A set $E$ is connected if and only if the only clopen subsets of $E$ are $\varnothing$ and the same set $E$. You can find the proof of this fact right here.