one point compactification I am asked to describe the one point compactification of $(0,1) \cup [2,3)$ of $\Bbb R$ and if I'm not mistaken it is just a circle union the closed set [2,3] correct? Am I missing something?
 A: A theorem specific to the one-point compactification: suppose you have a compact Hausdorff space $X$, then $X$ is (homeomorphic to) the one-point compactification of spaces of the form $X \setminus \{p\}$, where $p$ is a non-isolated point of $X$ (we need $X$ to be the closure of $X \setminus \{p\}$).
So e.g. if we remove a point (any point, as it is homogeneous) of the circle, we are left with an open interval essentially, so a space homeomorphic to $\mathbb{R}$. So the one-point compactification (which is essentially unique) of $\mathbb{R}$ is the circle. Similarly for the sphere and the plane.
In your case, if we have a circle $S^1$ and a disjoint compact segment $[2,4]$, say, we can say this is compact, but if we remove a point we have some cases: it's either the one-point compactification of $\mathbb{R}$ disjoint union a compact interval (if we remove a point from the circle), or of the circle union $[2,3) \cup (3,4]$, if we remove an interior point of the segment, or finally of $S^1 \cup [2,4)$, if we remove an endpoint. All of these are different from the space you started out with.
To see that Brian's answer (a circle with a tail, say $S^1 \cup ([1,2] \times \{0\}) \subset \mathbb{R}^2$) is correct, remove the point $(1,0)$ from that space and we are left with an open interval and a half-open segment, which is exactly your starting space.
A: You’re adding only one point: what you get is homeomorphic to the quotient of $[0,1]\cup[2,3]$ obtained by identifying $0,1$, and $3$ to a single point. It’s a circle with a tail, not the disjoint union of a circle and a segment.
