This is to finish flawed argument of Anonymous that $X$ has only zero-dimensional compactifications.

I'll prove that $K$ is totally disconnected, which is equivalent since $K$ is compact.

Let $S\subseteq K$ be connected and $|S|\geq 2$ where $K$ is a compactification of $X$. Since $\beta\omega\setminus \text{cl}_{\beta\omega} N$ is open and $\beta\omega$ is zero-dimensional, for $p\in S\cap \beta\omega\setminus \text{cl}_{\beta\omega} N$ and $q\in S, q\neq p$, one can find a clopen compact neighbourhood $U\subseteq \beta\omega\setminus \text{cl}_{\beta\omega} N$ of $p$ in $K$ such that $q\notin U$. So $S\subseteq \text{cl}_{\beta\omega} N\setminus N \cup K\setminus X$.

Let $f:\beta\omega \to K$ be the map induced by the injection $X\hookrightarrow K$ where $\beta\omega = \beta X$. Since any countable subset of $\beta\omega$ is $C^*$-embedded, we have that $\text{cl}_{\beta\omega} N =\beta N$ and so by restricting to it we can in fact just consider the map $g:\beta\omega\to L$ where $L = \text{cl}_K f(N) = f(\text{cl}_{\beta\omega} N)$, $S\subseteq L$, where $L$ is compact and $\beta\omega\setminus\omega$ is mapped by $g$ homeomorphically.

This implies that in particular $g(\omega) = L\setminus (\beta\omega\setminus \omega)$ is locally compact, open and zero-dimensional (being a countable Tychonoff space) in $L$, and so by the same argument as previously we can assume $S\subseteq \beta\omega\setminus \omega$.

But because $\beta\omega$ is zero-dimensional, we obtain a contradiction with existence of such set $S$.

And so $K$ must be zero-dimensional.