I'm trying to solve the following question:
Let $X$ be completely regular. Show that $X$ is connected if and only if $\beta(X)$ is connected. [Hint: If $X=A\cup B$ is a separation of $X$, let $f(x)=0$ if $x\in A$ and $f(x)=1$ if $x\in B$.]
My Attempt:
i) If $X$ is not connected, $\exists A, B$ open and disjoint s.t $X=A\cup B$. The function $f(x)=\cases{1, x\in A\\0, x\in B}$ is continuous. And, we can extend it to a function on $\hat{f}:\beta(X)\to [0,1]$. (Now, this is the part I'm unsure about:) $\beta(X)=\overline{X}=\overline{A}\cup \overline{B}$. And, $\hat{f}$ has to coincide with $f$ not only on $A$ and $B$, but in their closures since continuous extensions to closures are unique, and we have $\hat{f}(\{0\})\cup \hat{f}(\{1\})=\beta(X)$.
I'm not sure about this proof for a couple of reasons. First, not sure if $\overline{X}=\overline{A}\cup \overline{B}$ holds here though I feel like it should. Second, if that is indeed true, all the stuff done with the functions seems unecessary since $\overline{A}$ and $\overline{B}$ would be a separation of the compactification... On the other hand, the hint in the question lead me this way.
ii) I'm not sure how to proceed with this one. I feel like a separation of the compactification would imply a separation of the original set by intersecting the open sets that separate the former with the latter, but it doesn't seem to work out just like that, so I believe I'm missing something - although I do expect this direction to be easier than the other one.
(i): It's much easier if you treat $f$ as a function from $X$ to the two-point set $\{0,1\}$. It again has a continuous extension $\hat{f}$, and now it's clear that $\hat{f}^{-1}(\{0\}), \hat{f}^{-1}(\{1\})$ is a partition of $\beta X$ into two nonempty disjoint open sets.
(ii): Yeah, your idea should work. If you can write $\beta X = A \cup B$ where $A,B$ are disjoint, nonempty and open, consider $X \cap A$ and $X \cap B$. Clearly they are disjoint. To see they're open, recall that the inclusion $i : X \to \beta X$ is continuous (actually an embedding in this setting), so $i^{-1}(A) = X \cap A$ is open in $X$, and the same for $X \cap B$. (In other words, the subspace topology on $X \subset \beta X$ is at least as strong as the original topology; and under the completely regular assumption, the two topologies are equal.) Finally, to see they're nonempty, note that $X$ is dense in $\beta X$.