This is a follow-up to this question regarding one-space compactifications.
First recall a few definitions. An embedding is a continuous injective map $c:X\to Y$ that gives a homeomorphism from $X$ to its image. A compactification of $X$ is an embedding of $X$ as a dense subset of a compact space $Y$. By identifying $X$ with its image in $Y$ one just needs to extend $X$ to a bigger set $Y$ and give $Y$ a compact topology such that the subspace topology on $X$ coincides with the original topology on $X$.
Let $(X,\tau)$ be a noncompact topological space. A compactification $X^*=X\cup\{\infty\}$ obtained by adding a single point will be called a one-point compactification. (Wikipedia seems to reserve that term for the case of $X$ locally compact Hausdorff. In that case there is a unique Hausdorff one-point compactification of $X$. But I'll use the term more generally for convenience.)
I am interested in finding the range of all possible one-point compactifications of $X$. So find all the ways to give a compact topology to $X^*$ that induces the original topology on $X$. The fact that $X$ is dense in $X^*$ follows automatically because $X$ is not compact, so it cannot be closed in a compact space.
General fact: It is shown in the linked question at the top that every open nbhd of $\infty$ must be the complement in $X^*$ of a closed compact subset of $X$.
I can think of the following different cases of one-point compactifications:
The Alexandroff extension of $X$ with topology $$\tau_1=\tau\cup\{(X\setminus C)\cup\{\infty\}:C\text{ is compact and closed in }X\}\;.$$ Starting with the topology on $X$, we have added as many open nbhds of $\infty$ as possible. The inclusion map of $X$ into $X^*$ is an open embedding. This is the largest topology on $X^*$ with an open embedding.
The open extension topology given by $$\tau_2=\tau\cup\{X^*\}\;.$$ Starting with the topology on $X$, we have added a single nbhd of $\infty$, namely the whole space. The inclusion map is an open embedding. This is the smallest topology on $X^*$ with an open embedding.
Any topology intermediate between $\tau_1$ and $\tau_2$. $X$ will also be open in $X^*$ and the topology will contain all of $\tau$. For nhbds of $\infty$ one has to pick of suitable subfamily of the complements of all closed compact subsets of $X$. For example, the complements of all finite closed compact subsets, or the complements of all countable closed compact subsets.
These should cover all cases of one-point compactification with an open embedding. ($X$ is open in $X^*$ if and only if the topology on $X^*$ contains all of $\tau$.)
Now is every one-point compactfication of $X$ always an open embedding? Are there examples where $X$ is not open in $X^*$? What are conditions on $X$ that ensure the only one-point compactifications are the ones above?
Side note: One is usually not interested in taking a one-point compactification of a space $X$ that is already compact. But I still find the following observation worth noting. If $X$ is compact, the Alexandroff extension of $X$ will not be a compactification, since $\infty$ is then an isolated point and hence $X$ is not dense in $X^*$. But the open extension topology on $X$ plus one point is always a one-point compactification (with $X$ open in $X^*$).
If $X$ is a one-point space, it has exactly two one-point compactifications. The open extension topology is the Sierpinski space. The other is the two point space with the indiscrete topology, in which case the embedding is not open.
If $X$ is not compact, then every one-point compactification of $X$ is an open embedding. Indeed, suppose $X$ is not open in a one-point compactification $X^*=X\cup\{\infty\}$. This means there is some $x\in X$ such that every open neighborhood of $x$ in $X^*$ contains $\infty$. But then every open neighborhood of $x$ contains a neighborhood of $\infty$, and thus its complement is a compact subset of $X$. Since $X$ is embedded in $X^*$, this means that every open neighborhood of $x$ in $X$ has compact complement. But this implies $X$ is compact, since every open cover of $X$ includes a set which contains $x$, and then the complement of that set is covered by finitely many other sets in the cover.