Exercise on geometrically connectedness of a scheme I want to prove the following exercise from Götz' & Wedhorn's Algebraic Geometry I on page 144: Chapter 5: Schemes over fields (only part (a); part (b) is a simple consequence):

Exercise 5.23. Let $k$ be a field and let $X$ be a connected $k$-scheme.
  (a) Assume that there exists a non-empty geometrically connected $k$-scheme $Y$ and a
  $k$-morphism $f:Y \to X$. Show that $X$ is geometrically connected.
Hint: Use Exercise 5.22 to show that $X_{\Omega} \to X$ is open, closed, and surjective for
  a separable closure $\Omega$ of $k$ and show that there are no nontrivial open and closed
  subsets of $X_{\Omega}$.

Q: I haven't found the key to show the last part of the exercise that there are no nontrivial open and closed subsets in $X_{\Omega}$. 
By proposition 5.53 $X$ is geometrically connected if and only if $X_K= X \times_k Spec \text{ } K$ is connected for certain separable closed field extension 
$k \to K$. thus it suffice to show that $X_{\Omega}$ is connected. because $Y$ is gemetrically connected, the pullback $Y_{\Omega}$ is connected. in addition I partially succeed in proving 5.22 except of the $p:X_{\Omega} \to X$ is open part (???). does it follow from the 'closed' part? I don't see it.
if we continue with the main claim to show that $X_{\Omega}$ is connected then let assume that $X_{\Omega}$ is not connected and we can find a non empty open & closed proper subset $U \subsetneqq X_{\Omega}$. then $p(U)$ is also non empty, open & closed in $X$ and therefore $p(U)= X$ because $X$ connected and $p$ open& closed. Similary  $p(X_{\Omega} \backslash U)= X$. 
in addition since $Y_{\Omega}$ is connected one the preimages of $U$ or $X_{\Omega} \backslash U$ under induced map $f_{\Omega}: Y_{\Omega} \to X_{\Omega}$ is empty and the other the whole $Y_{\Omega}$.  at this point I'm a bit lost. can I obtain from this observations a contradiction on non emptyness of $U$?
 A: The hint refers to 5.22:

Exercise 5.22. Let $A\to B$ be an integral ring homomorphism. Show that the corresponding morphism of schemes $\operatorname{Spec} B\to \operatorname{Spec} A$ is universally closed. Deduce that if $K$ is an algebraic extension of a field $k$, then the projection $X_K\to X$ is universally open, universally closed, and surjective.

You mention you have not yet proven that $X_K\to X$ is universally open. If we prove this, then we have that $X_\Omega\to X$ is universally open, universally closed and surjective, and then we're in the situation of your other post that I responded to, where the appropriate fiber product diagram is 
$\require{AMScd}$
\begin{CD}
Y_\Omega @>p_X>> X_\Omega\\
@Vp_YVV @VVfV\\
Y @>g >> X
\end{CD}
where all the schemes besides $X_\Omega$ are connected by assumption. Thus $X_\Omega$ must also be connected by the result at that question.
In order to prove that $X_\omega\to X$ is universally open, it is enough to prove that $\operatorname{Spec} K\to \operatorname{Spec} k$ is universally open for an algebraic field extension $k\subset K$. By a standard locality argument, it's enough to check this on affines. So we want to check that for $R$ any $k$-module, the map $R\to R\otimes_k K$ is open. It suffices to prove that the image of a standard open $D(f)$ is again open for $f\in R\otimes_k K$. In order to do this, pick some finite algebraic $k$-subfield of $F\subset K$ so that $f\in R\otimes_kF$. Then $R\to R\otimes_kF$ is flat and finitely presented, so the image of $D(f)$ from this map is open, and it is the same image as we get when considering $D(f)$ coming from $R\otimes_kK$.
To prove that the images are the same, consider the obvious commutative triangle of schemes $\operatorname{Spec}(R\otimes_k K)_f \stackrel{\alpha}{\to} \operatorname{Spec}(R\otimes_k F)_f \stackrel{\beta}{\to} \operatorname{Spec} R$ where the third side $\beta\circ\alpha$ isn't drawn since diagrams here are difficult. Suppose a prime ideal $p\subset R$ is in the image of $\beta$. This means that $(R\otimes_k F)_f\otimes_R \kappa(p) = (F\otimes_k \kappa(p))_f$ is nonzero, so $F\otimes_k\kappa(p)$ is nonzero and the image of $f$ in it is not nilpotent. Next, the ring map $F\otimes_k \kappa(p)\to K\otimes_k \kappa(p)$ is injective, so $(R\otimes_k K)\otimes_R\kappa(p) = K\otimes_k \kappa(p)$ is not the zero ring and the image of $f$ in it is not nilpotent. So $(R\otimes_k K)\otimes_R \kappa(p)$ isn't the zero ring and we've shown what we wanted.
