My attempt at specifying a homeomorphism is as follows.
Let $(X,\tau), (X\times[0,1]/\sim, \tau')$ and $(X\times[0,1),\tau'')$ be topological spaces. Let p be the equivalence class of $X\times\{1\}$ in $X\times[0,1]/\sim$. Let $\pi$ be the quotient map from $X \times [0,1]$ to $(X\times[0,1]/\sim, \tau')$.
Let $(\chi^+,\tau''^+)$ be the one-point compactification of $X\times[0,1)$, i.e. $\chi^+ := (X\times[0,1))\cup\{\infty\}$.
Let $f: (X\times[0,1]/\sim, \tau') \to (\chi^+,\tau''^+)$ be defined as:
$$ f(x) =
\begin{cases}
x, & \text{if $x \in X\times[0,1)$} \\
\infty, & \text{if $x =p$}
\end{cases} $$
I claim that f is the homeomorphism that we are after.
That f is bijective is easy to see. Now I want to show that f is continuous.
Let $U \in \tau''^+$. We want to show that $f^{-1}(U)$ is open. If $\{\infty\} \notin U$, then $U$ is already in $\tau'$ and we are done.
Otherwise, let $U = ((X\times[0,1)) \setminus C) \cup \{\infty\}$, where $C \subset X\times[0,1)$ is a closed, compact set.
We have: $f^{-1}(U) = f^{-1}(((X\times[0,1))\setminus C) \cup \{p\} = ((X\times[0,1))\setminus C) \cup \{p\} $. And so $\pi^{-1}(f^{-1}(U)) = \pi^{-1}(((X\times[0,1))\setminus C) \cup \{p\}) = ((X\times[0,1))\setminus C) \cup X\times\{1\} = X\times[0,1]\setminus C$.
Since C is closed, we have that $X\times[0,1]\setminus C$ is open in $X\times[0,1]$. So $((X\times[0,1))\setminus C) \cup \{p\}$ is open in the quotient topology of $(X\times [0,1])/\sim$, i.e. $f^{-1}(U)$ is open in $(X\times [0,1])/\sim$. This proves the continuity of f.
Before proceeding with the continuity of $f^{-1}$, I'd like to ask for others' opinion on the part above. Also, just to be sure, in $X\times [0,1]$, are we considering $[0,1]$ as a subset of $\mathbb{R}$, i.e. does it has the subspace topology induced by the usual topology on $\mathbb{R}$? If not what is the topology on $[0,1]$?
$C(X)$ minus the identified point $\{b\}$ (the identified $X \times \{1\}$ set) is just homeomorphic to $X \times [0,1)$ as $p$ restricted to that set is 1-1 and open and continuous.
Now apply the theorem that if $Y$ is compact Hausdorff and for some $y \in Y$, $Y\setminus \{y\} \simeq X$, then $Y$ is homeomorphic to the one-point compactification of $X$, and note that $C(X)$ now applies.