Let $X$ be a completely regular space (that is, any closed subset and point not in the closed set can be separated by continuous function).
Let $C = C(X,[0,1])$ be the collection of continuous functions $f$ defined from $X$ to $[0,1].$
Recall the construction of Stone-Cech Compactification of $X$ via embedding in a cube.
For any $x\in X,$ defined a map $\phi:X\to [0,1]^C$ by $\phi(x) = (f(x))_{f\in C}.$ By Tychonoff's theorem, the space $[0,1]^C$ is compact in Tychonoff product topology. Note that $\overline{\phi(X)},$ which is a closed subset of a compact set, is compact in $[0,1]^C$. Then we let $\overline{\phi(X)}$ to be the Stone-Cech compactification of $X,$ denoted as $\beta X.$
Question: Let $V$ be an open subset in $\beta X.$ Is it true that $U = V\cap X$ open in $\beta X?$
Firstly, the intersection is not well-defined as it stands. $X$ is not a subset of $\beta X$ in this setup. What does hold is that $\phi[X]$ is a subset of $\beta X \subseteq [0,1]^C$ which is homeomorphic to $X$ (this is what $\phi$ being an embedding means). Often $X$ is identified with this image, also in notation. So $V$ open in $\beta X$ (so relatively open in the product) means that $V \cap \phi[X]$ is open in $\phi[X]$ (or $X$), not in $\beta X$, unless $X$ or $\phi[X]$ is open in $\beta X$ (which happens iff $X$ is locally compact Hausdorff).
So making the identification $X \sim \phi[X]$ : for all open $V \subseteq \beta X$: $V \cap X$ open in $\beta X$ iff $X$ is locally compact Hausdorff.