The Stacks project

Proof. Write $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open and closed. Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base change of $f$. Since $T_{\overline{k}}$ is connected we see that $T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$ or $\overline{f}^{-1}(\overline{V})$. Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$.

Fix a quasi-compact open $W \subset X$. There exists a finite Galois subextension $\overline{k}/k'/k$ such that $\overline{U} \cap W_{\overline{k}}$ and $\overline{V} \cap W_{\overline{k}}$ come from quasi-compact opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \amalg V'$. Consider

These are Galois invariant, open and closed, and $W_{k'} = U'' \amalg V''$. By Lemma 33.7.10 we get open and closed subsets $U_ W, V_ W \subset W$ such that $U'' = (U_ W)_{k'}$, $V'' = (V_ W)_{k'}$ and $W = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W \cap U_{W'} = U_ W$ and $W \cap V_{W'} = V_ W$. Verification omitted. Hence we see that upon defining $U = \bigcup _{W \subset X} U_ W$ and $V = \bigcup _{W \subset X} V_ W$ we obtain $X = U \amalg V$. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $f(T)$ by construction. $\square$