If the Stone-Čech compactification $\beta X$ of $X$ is connected then $X$ is connected So I am searching to prove that if the Stone-Čech compactification $\beta X$ of a Tychonoff space $X$ is connected then $X$ is connected. Therefore, I tried to prove that $\beta[X]$ is connected and in particular I tried to procced as by reduction ad absurdum. So if $\beta[X]$ is not connected then there exist two disjoint set $A_1$ and $A_2$ of $\beta X$ such that $\beta[X]$ is disjoint union of $A_1\cap \beta[X]$ and $A_2\cap \beta[X]$. So I observed that
$$
\beta[X]\subseteq A_1\cup A_2
$$
and thus I conclude that $A_1\cup A_2$ is dense so that $X$ is union of $\operatorname{cl}A_1$ and $\operatorname{cl}A_2$ but this does not seem help; moreover I observed that
$$
(A_1\cap A_2)\cap\beta[X]=\emptyset
$$
so that by density of $X$ I clonclude that $A_1$ and $A_2$ are disjoint and thus
$$A_1\subseteq\beta X\setminus\operatorname{cl}A_2\quad\text{and}\quad A_2\subseteq\beta X\setminus\operatorname{cl} A_1
$$
and thus I conclude that $X$ is union of $X\setminus A_1$ and $X\setminus A_2$ so that $(X\setminus A_1)\cup (X\setminus A_2)$ is dense but another time this does not seem help.
Anyway, I know that a topological space $T$ is connected if and only if there not exists continuous function form $T$ to the discrete space $\{0,1\}$ so that I tried to define a function
$$
f(x):=\begin{cases}0,\,\text{if }x\in A_1\cap\beta[X]\\
1,\,\text{if }x\in A_2\cap\beta[X]\end{cases}
$$
for any $x\in\beta[X]$ and thus I tried to extend it to a continuous function on $\beta X$ but I unfortunately failed: however here is said that this is a trivial fact but I really do not see this.
Finally, I tried to use the following hint given by my topology text.

Hint: the clousure of a clopen set on $\beta[X]$ is clopen in $\beta X$.

However I was not able to use this hint and thus I thought to put a specific question. So could someone help me, please?
 A: A much less sophisticated argument:

If $X$ is disconnected, there exists a non-constant continuous map $\varphi:X\to K$ where $K=\{0,1\}$ with the topology $\{\emptyset,\{0\},\{1\},\{0,1\}\}$. $K$ is trivially a compact Hausdorff space: by the universal property of the Stone-Cech compactification, there must exist a continuous map $\psi:\beta X\to K$ (forced to be non-constant by the conditions of the universal property) which forces $\beta X$ to be disconnected.

A: If $X$ is not connected, there are nonempty open $U,V$ which disconnect $X$. $\varphi:X\to[0,1]$, $\varphi|_U\equiv0$ and $\varphi|_V\equiv1$ is continuous.
I use the construction of:

Let $\eta_X:X\to[0,1]^{C(X,[0,1])}$, $x\mapsto(f(x))_{f\in C(X,[0,1])}$, where $[0,1]^{C(X,[0,1])}$ has the product topology. Let $\beta X:=\overline{\eta_X(X)}$ with the subspace topology, and $\pi_f:\beta X\to[0,1]$ the projection maps for each $f\in C(X,[0,1])$.

Then in particular, $\mathcal{O}_1:=\beta X\cap\pi_{\varphi}^{-1}\{[0,1/2)\}$ and $\mathcal{O}_2:=\beta X\cap\pi_{\varphi}^{-1}\{(1/2,0]\}$ are both open in $\beta X$. To check whether or not $\mathcal{O}_1\sqcup\mathcal{O}_2=\beta X$ we need only check this on $\pi_{\varphi}(\beta X)$. It can be shown that each $x$ in this set must lie in the closure of $\varphi(X)$, which is $\varphi(X)=\{0\}\sqcup\{1\}$ itself. Therefore $\mathcal{O}_1\cap\mathcal{O}_2=\emptyset$ (and each set is obviously nonempty) and each element $\zeta$ of $\beta X$ is in exactly one of these sets since $\pi_{\varphi}(\zeta)\in\overline{\varphi(X)}=\{0,1\}$ must hold, and these points are covered by the $\mathcal{O}_{1,2}$.
This shows that $X$ disconnected implies $\beta X$ disconnected (no further hypotheses on the topology of $X$), so $\beta X$ connected must imply $X$ is connected.
If you want me to expand on why each $x\in\pi_{\varphi}(\beta X)$ must lie in $\overline{\varphi(X)}$, I will add this.
Addition:

In this self-answer I prove that for all topological spaces $X$, all compact Hausdorff spaces $K$ and all continuous $\varphi:X\to K$, we have a unique continuous $\psi:\beta X\to K$ with $\psi\circ\eta_X\equiv\varphi$ (universal property) and the related fact that we must have $\pi_{h\circ\varphi}(b)\in h^{-1}\overline{\varphi(X)}$ for all $b\in\beta X$ and all $h\in C(K,[0,1])$.
As $K=[0,1]$ is the space I took here, with $\varphi$ being the constant restrictions on $U,V$ in the rest of this post, I may let $h=\mathrm{id}_K$ and claim that for all $b\in\beta X$, $\pi_{\mathrm{id}_K\circ\varphi}(b)\in(\mathrm{id}_K)^{-1}\overline{\varphi(X)}$, that is, $\pi_{\varphi}(b)\in\overline{\varphi(X)}$ must hold.

I add the relevant portion of the answer here for convenience:

Fix $b\in\beta X=\overline{\eta_X(X)}$, and take any continuous $h:K\to[0,1]$. There are different ways to go about the following, but we may as well use a sequence as that is the "nice" way. For all $n\in\Bbb N$ let $F_n=[b(h\circ\varphi)-1/n,b(h\circ\varphi)+1/n]\cap[0,1]$. As $h$ is continuous, $h^{-1}\{F_n\}$ is closed and thus compact in $K$; moreover, $h^{-1}\{F_{n+1}\}\subseteq h_0^{-1}\{F_n\}$ always.
By the product topology, we can take a basic, cylinder-set neighbourhood of $b$ on the coordinate $h\circ\varphi$ and the nonempty open set $F_n^{\circ}$ for each $n$. As $\eta_X(X)$ is dense in $\beta X$, this neighbourhood of $b$ will contain some $\eta_X(x_n)$ for some $x_n\in X$, thus $\eta_X(x_n)(h\circ\varphi)=h(\varphi(x_n))\in F_n^{\circ}$.
That guarantees $h^{-1}\{F_n\}$ is never empty, as it contains $\varphi(x_n)$. By the finite intersection property and compactness, $A_h:=\bigcap_{n\in\Bbb N}h^{-1}\{F_n\}$ is a nonempty compact subset of $K$. Each element $a\in A_h$ is such that $h(a)$ is in all the $F_n$; by basic properties of $[0,1]$ we must then have that $h(a)=b(h\circ\varphi)$ for all $a\in A_h$. We can also note that $a\in\overline{\varphi(X)}$.
Take these nonempty compact $A_h$ for all continuous $h:K\to[0,1]$. We want to show that their intersection is nonempty, since the intersection is necessarily a one-point set (if it is nonempty) by the Urysohn lemma. We can assert a nonempty intersection if we demonstrate the finite intersection property. Let $\{h_i\}_{i=1}^k\subset C(K,[0,1])$ be arbitrary - we want to show $\bigcap_{i=1}^kA_{h_i}\neq\emptyset$. Let, for $1\le i\le k$ and $n\in\Bbb N$, $F^{(i)}_{n}:=[b(h_i\circ\varphi)-1/n,b(h_i\circ\varphi)+1/n]\cap[0,1]$. Similarly can we construct basic cylinder neighbourhoods of $b$, for fixed $n$, by asserting the neighbourhood $(F^{(i)}_n)^{\circ}$ on the coordinates $h_i\circ\varphi$. By density will this contain an $\eta_X(x_n)$ for some $x_n\in X$, with $h_i(\varphi(x_n))\in(F^{(i)}_n)^{\circ}$ for all $i$. Then the intersected preimage $B_n:=\bigcap_{i=1}^kh_i^{-1}\{F^{(i)}_n\}$ is nonempty (and compact by continuity of each $h_i$) for each $n$, with the same property $B_{n+1}\subseteq B_n$ for all $n$. We have by the finite intersection property: $$\emptyset\neq\bigcap_{n\in\Bbb N}B_n\subseteq\bigcap_{i=1}^kA_{h_i}$$Which concludes the proof: $$\exists!a_b\in\bigcap_{h\in C(K,[0,1])}h^{-1}b(h\circ\varphi),\,\forall b\in\beta X$$It is pleasing to observe that this $a_b$ is in the closure $\overline{\varphi(X)}$ always, and can be intuitively thought of as a limit point of the $\eta_X(x)$ approaching $b$.

