Connected scheme with a section is geometrically connected Greeting, I'm trying to resolve the following exercise using the hint below

So far I managed to prove that $\mathrm{Spec}\, k' \rightarrow \mathrm{Spec}\, k$ is both universally open and universally closed. Hence $p:X_{k'}\rightarrow X$ is open and closed, moreover it's finite and flat as well, thus making $X_{k'}$ into a locally noetherian space. Its connected component are therefore open and closed i.e. $p(U)=X$ for a connected component $U$ of $X_{k'}$. Still i can't see why the base change $x_{k'}$ of a rational point must meet $U$.
Thanks in advance.
 A: Here's what you already did: If $X_{k'}\to X$ is open and closed, that means any connected component of $X_{k'}$ surjects on to a connected component of $X$: the open+closed immersion of the connected component $Z\to X_{k'}$ composes with the open+closed map $X_{k'}\to X$, so that $Z\to X$ is open and closed, and thus the image of $Z$ is both open and closed, and thus a connected component.
To finish, if $X$ is connected, then every connected component surjects on to $X$, which means all the connected components have the same $k$-rational point $x\in X$ in their image. But the base change of this rational point along $\operatorname{Spec} k'\to \operatorname{Spec} k$ is just a single $k'$-rational point, so all the connected components of $X_{k'}$ meet at this single $k'$-rational point, which means that $X_{k'}$ must be connected. It might help to consider the following diagram:
$$\require{AMScd}
\begin{CD}
\operatorname{Spec} k' @>{}>> X_{k'} @>{}>> \operatorname{Spec} k'\\
@VVV @VVV @VVV\\
\operatorname{Spec} k @>{}>> X @>>> \operatorname{Spec} k
\end{CD}$$
The bottom is the inclusion of $\operatorname{Spec} k$ into $X$ hitting the $k$-rational point $x$ followed by the projection to $\operatorname{Spec} k$, so that the composite is the identity. Thus the top row must have its composite be the identity, too, and the first map in the top row picks out the unique base change of our rational point to $k'$. But there's only one of these, and it has to be in every connected component, so there can only be one connected component of $X_{k'}$, so it's connected.
