One-point compactification of $(0,1) \cup (2,3)$ Definition from Munkres, the textbook we are following: If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y \setminus X$ equals a single point then, then $Y$ is called the one-point compactification of $X$. From this definition, I understood it this way: I just need to add one single point to the given space and it would be compact and Hausdorff. But don’t I need to add 4 points? Namely 0, 1, 2 and 3. Because these are the limit points that the given space lacks to be closed, and in return, compact. What am I missing?..
 A: You are missing that you have to consider the intervals in isolation, not as subsets of $\mathbb R$. That means there's no interval $[1,2]$ sitting in between, and also no intervals on either side. And you may move them around and bend them as much as you like (as long as you don't tear them apart).
In particular, you can bend them around to get an 8-like figure with only the crossing point missing. The one-point compactification then adds exactly that crossing point.
A more methodical way would be to first compactify the set in any way you want (with the only restriction that your original set must be open in the compactification), and then to identify all the points you added with each other. In your example, you could use your four-point compactification, and then identify the four points with each other.
The ultimate methodical way is, of course, to just apply the definition of the one-point compactification, and then think about what the resulting space looks like.
A: Look at $$X=\{(x,y)\in \Bbb R^2\mid (x+1)^2 + y^2= 1\} \cup \{(x,y)\in \Bbb R^2\mid (x-1)^2 + y^2= 1\}$$
which is the subspace of the plane of two circles intersecting exactly at $(0,0)$.
$X$ is compact as the union of two compact unit circles. (or closed and bounded whatever you like).
$X\setminus \{(0,0)\}$ consists of two disjoint open intervals, topologically. Any circle minus a point is just homeomorphic to $(0,1)$ (e.g for the unit circle around the origin minus $(1,0)$ just use the homeomorphism $(0,1) \ni t \to (\cos(2\pi t), \sin(2 \pi t))$ but all points behave the same on all circles so this holds for our two circles of $X$ minus $(0,0)$ too).
And two disjoint open is just our original space (the gap between them in $\Bbb R$ is irrelevant, all that matters is that they're two disjoint open intervals).
So $X$ is the one-point compactification of $(0,1) \cup (2,3)$ : it's compact and there is one point (the compactifying point as it were) that we can remove and get our original space back (two disjoint open intervals, up to homeomorphism).
This is how you can tell that $X$ is the (unique up to homeomoerphism!) one-point compactification. You knwo it exists because $(0,1) \cup (2,3)$ is locally compact and Hausdorff.
A: Think about just one interval $(0,2\pi)$ first. If you add 2 points, you could make the compact $[0,2\pi]$. But the one point compactification does it differently. It only adds one point so that the space becomes a circle like $(0,2\pi)$ together with $0 \equiv 2\pi$ are angles around the unit circle.
With 2 intervals, you could add 4 points as you are thinking and get something compact. But one point compactification does something different. The neighborhoods of the added point are totally different.
