**Proof.**
Let $\overline{T}$ be an irreducible component of $X_{\overline{k}}$. We may choose an affine open $U \subset X$ such that $\overline{T} \cap U_{\overline{k}}$ is not empty. Write $U = \mathop{\mathrm{Spec}}(A)$, so $A$ is a finite type $k$-algebra, see Morphisms, Lemma 29.15.2. Hence $A_{\overline{k}}$ is a finite type $\overline{k}$-algebra, and in particular Noetherian. Let $\mathfrak p = (f_1, \ldots , f_ n)$ be the prime ideal corresponding to $\overline{T} \cap U_{\overline{k}}$. Since $A_{\overline{k}} = A \otimes _ k \overline{k}$ we see that there exists a finite subextension $\overline{k}/k'/k$ such that each $f_ i \in A_{k'}$. It is clear that $\text{Gal}(\overline{k}/k')$ fixes $\overline{T}$, which proves (1).

Part (2) follows by applying Lemma 33.8.12 (1) to the situation over $k'$ which implies the irreducible component $\overline{T}$ is of the form $T'_{\overline{k}}$ for some irreducible $T' \subset X_{k'}$.

To prove (3), let $T \subset X$ be an irreducible component. Choose an irreducible component $\overline{T} \subset X_{\overline{k}}$ which maps to $T$, see Lemma 33.8.10. By the above the orbit of $\overline{T}$ is finite, say it is $\overline{T}_1, \ldots , \overline{T}_ n$. Then $\overline{T}_1 \cup \ldots \cup \overline{T}_ n$ is a $\text{Gal}(\overline{k}/k)$-invariant closed subset of $X_{\overline{k}}$ hence of the form $W_{\overline{k}}$ for some $W \subset X$ closed by Lemma 33.7.10. Clearly $W = T$ and we win.
$\square$

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