One-point compactification problem

Thank You all for your answers. I now write my answer and hope someone would give me some feedback.

A=[0,1)U[2,3). Then I chose f:[0,2]->A+. Want to show that the map is bijectiv and continous.

Bijectiv because:

f:[0,2]->A+

[0,1)->[0,1) (1,2]-> [2,3) 1-> x_0 where x_0 is the point at infinity in the one-point compactification A+ of A. Is it right?

Continous: Take U open in [0,2] and want to show that the image is open in A+. But from here I am confused?

Theorem:

If $X$ is a locally compact Hausdorff space and $Y$ is compact Hausdorff such that for some point $p \in Y$, $X$ is homeomorphic to $Y\setminus \{p\}$, then $Y$ is homeomorphic to the one-point compactification of $X$.

Proof: Let $h: X \to Y\setminus \{p\}$ be the promised homeomorphism. Define $h': \alpha(X) = X \cup \{\infty\} \to Y$ by $h'(x) = h(x)$ for $x \in X$ and $h'(\infty) = p$. Then $h'$ is clearly a bijection. To see it is continuous, let $O \subseteq Y$ be open. If $p \notin O$, then $O \subseteq Y\setminus \{p\}$ and this set is open in that subspace so that $h'^{-1}[O] = h^{-1}[O]$ is open in $X$ and so open in $\alpha(X)$. If $p \in O$ then $Y \setminus O \subseteq Y\setminus \{p\}$ is compact and so is $C:= h^{-1}[Y \setminus O]$ as $h$ is a homeomorphism, and $h'^{-1}[O] = \{\infty\} \cup (X\setminus C)$ is also open in $\alpha(X)$ (by the definition of the topology on the one-point compactification $\alpha(X)$). So $h'$ is a continuous bijection from a compact space to a Hausdorff space hence a homeomorphism.

Then note that $(1,2] \simeq [2,3)$ (via $f:[2,3) \to (1,2]; f(x) = -x + 4$) and so

$$[0,1) \cup [2,3) \simeq [0,1) \cup (1,2] = [0,2] \setminus \{1\}$$

so $[0,2]$ is homeomorphic to the one-point compactification of the first space.