Is quotient of open invariant subset open?

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I am reading GIT book by Mumford. He needs special cases of the following conjecture several times.

Conjecture Let $G$ be a reductive algebraic group acting on an irreducible affine scheme $X=Spec A$. Let $Y = Spec A^G$ and $\phi: X \rightarrow Y$ is canonical map. Let $U$ be an open invariant subset in $X$. Then $\phi (U)$ is open.

Question Is it true?

Remark 1. It is a theorem from Mumford’s book that $\phi$ is a categorical quotient.

Remark 2 As far as Mumford do not formulate the conjecture as a proposition in his book, I expect it to be false.

Edited My question is closely related to this one. user31480 has asked about invariant subsets, I have asked about any. I have just realized, that it is not essential at all. For any open subset $U$ one can consider union of $gU$ for all $g \in G$ (still open, has the same image by quotient map, but invariant).

Also I noticed that assumption of $X$ being irreducible is crucial. You can find a counterexample as an answer for aforementioned question.