A topological space $X$ is totally disconnected if the connected components in $X$ are the one-point sets, and a topological space, $X$ is called completely regular exactly in case points can be separated from closed sets via continuous real-valued functions. Let $X$ be a totally disconnected and completely regular topological space. Can we deducd that $\beta X$ is also totally disconnected, where $\beta X$ is the Stone–Čech compactification of $X$?
No; $X$ may have a quasi-component with more than one point, and each quasi-component of $X$ is contained in a connected subset of $\beta X$. It's easy to construct examples in $\mathbb R ^2$.
Even if the quasi-components of $X$ are trivial, $\beta X$ may have a connected subset with more than one point. The Erdös space $\mathfrak E$ has singleton quasi-components, but is not zero-dimensional, therefore $\beta \mathfrak E$ is not zero-dimensional, equivalently, $\beta \mathfrak E$ is not totally disconnected.
Even if $X$ is zero-dimensional, $\beta X$ may have nondegenerate connected subsets. There is no separable metrizable $X$ to this effect, but there does exist a Tychonoff (completely regular T$_1$) example.