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 in $\beta\omega$ and $\beta\omega$ is zero-dimensional, we must have $S\subseteq \text{cl}_{\beta\omega} N\setminus N \cup K\setminus X$ (for if $p\in \beta\omega\setminus \text{cl}_{\beta\omega} N$ and $q\neq p$ then one can take a clopen compact $U\subseteq\beta\omega\setminus \text{cl}_{\beta\omega} N$ with $p\in U$ and $q\notin U$, the same argument applies in the next claims but I won't repeat it).

Since any countable subset of $\beta\omega$ is $C^*$-embedded, we have that $\text{cl}_{\beta\omega} N =\beta N \cong \beta\omega$ and $\beta\omega\setminus \omega\cong \text{cl}_{\beta\omega} N\setminus N$. Consider the compact set $L = \text{cl}_{\beta\omega} N\setminus N \cup K\setminus X$.

The injection $X\hookrightarrow K$ induces a map $\beta X = \beta\omega\to K$ mapping $\beta\omega\setminus X = N$ to $K\setminus X$, so that $K\setminus X$ is countable.

This implies that in particular $K\setminus X$ is open and zero-dimensional in $L$, and so we can assume $S\subseteq \text{cl}_{\beta\omega} N\setminus N$.

But because $\beta\omega\setminus \omega\cong \text{cl}_{\beta\omega} N\setminus N$ is zero-dimensional, we obtain a contradiction.

And so $K$ must be zero-dimensional.