Continuity of a function from a one-point compactification to a cone Let $X$ a compact Hausdorff space and let the cone 
\begin{align}
CX := (X \,  \times \, [0, 1]) \, / \, A
\end{align}
where $A:= X \times \{1\} \subset X \, \times \, [0, 1]$. Show that $CX$ is homeomorphic to the one-point compactification $Y^{+}$ of $Y$, where $Y:= X \times [0, 1)$ and $Y^{+}:= Y \overset{\cdot}\cup \{ p\} $.
I tried to follow the answer at the question on this topic (Homeomorphism between a cone in a compact space and this compactification).
I have some trouble to prove that the function $h: Y^{+} \rightarrow CX$ definied by \begin{align}h(y) : = \begin{cases} f(y), \, \text{if} \, \,  y \in Y \\
a, \, \text{if} \, \, y = p
\end{cases} \end{align}
is continuous in the case of the point $a\in U$, where $U \subset CX$ open set, $f: X \times [0, 1] \rightarrow CX= (X \times [0, 1])\,  / \, A$ is the quotient map and $a \in CX$ is the point corresponding to $A$ in $X \times [0, 1]$. 
I want to show that the preimage $h^{-1}(U) \subset Y^{+}$ is open. By definition  I know that a set $V \subset Y^{+}$ is said to be open if and only if \begin{cases} V \subset X \, \, \text{open in} \, \, X, \text{if} \, p \notin V\\ V = Y^{+} \setminus
 C \, \, \text{for a compact set} \, \, C\subset X, \text{if} \, \, p \in V\end{cases}
but I can't see how use it to prove the continuity of $h$ in the case where the point $a \in U$.
Any suggestions? Thanks in advance!
 A: 
For an open set $U\subset CX$ containing $a=f(A)$, show that the preimage $h^{-1}(U)\subset Y^+$ is open.

Since $h(p)=a\in U$, $p\in h^{-1}(U)$. By the definition of the one-point compactification, it suffices to find a compact set $C\subset Y$ such that $h^{-1}(U)=Y^+\setminus C$.
We claim $C\equiv Y\setminus f^{-1}(U)=(X\times[0,1])\setminus f^{-1}(U)$. 
(The second equality comes from the assumption $h(p)=a\in U \Rightarrow f^{-1}(a)=X\times\{1\}\subset f^{-1}(U)$.)


*

*Since the quotient map $f$ is continuous, $f^{-1}(U)$ is open in $X\times[0,1]$ and so $C$ is closed. Furthermore, since $C$ is a closed subset of a compact space $X\times[0,1]$, $C$ is also compact.

*It remains to show that $h^{-1}(U)=Y^+\setminus C$. Note that $p\in h^{-1}(U)$ by assumption and
$$
p\notin Y \Rightarrow p\notin Y\setminus f^{-1}(U)\equiv C \Rightarrow p\in Y^+\setminus C
$$
Thus it suffices to show $h^{-1}(U)\setminus\{p\}=Y\setminus C=f^{-1}(U)$. But it is trivial since $h(x)=f(x)$ for all $x\in Y=X\times[0,1)$.
