We have two spaces: $[0,1]/C$ and $[0,1]/A$, where $C$ denotes the Cantor set and $A=\{0,1,\frac{1}{2},\frac{1}{3},...\}$. One needs to show they are homeomorphic.
What I thought about is showing $A\cong C$, but I suppose it is not true... What should I do? Show the same sets are open in both spaces?
Let $X_C=[0,1]/C$ and $X_A=[0,1]/A$. Let $p_C$ be the point of $X_C$ corresponding to $C$, and let $p_A$ be the point of $X_A$ corresponding to $A$. $[0,1]\setminus C$ is the union of a countably infinite family $\mathscr{I}_C$ of pairwise disjoint open intervals, and $X_C\setminus\{p_C\}$ is clearly homeomorphic to $[0,1]\setminus C=\bigcup\mathscr{I}_C$. Similarly, $[0,1]\setminus A$ is the union of a countably infinite family $\mathscr{I}_A$ of pairwise disjoint open intervals, and $X_A\setminus\{p_A\}$ is homeomorphic to $[0,1]\setminus A=\bigcup\mathscr{I}_A$. Thus, the subspaces $X_C\setminus\{p_C\}$ and $X_A\setminus\{p_A\}$ are homeomorphic to $(0,1)\times\Bbb N$ and hence to each other.
Let $\pi_C:[0,1]\to X_C$ and $\pi_A:[0,1]\to X_A$ be the quotient maps.
$U\subseteq X_C$ is an open nbhd of $p_C$ iff $\pi_C^{-1}[U]$ is an open nbhd of $C$ in $[0,1]$. Let $V=\pi_C^{-1}[U]$ be an open nbhd of $C$, and let $K=[0,1]\setminus V$. $C$ and $K$ are disjoint closed sets, so
$$d=\inf\{|x-y|:x\in C\text{ and }y\in K\}>0\;.$$
It follows that if $I\in\mathscr{I}_C$ has length less than $d$, then $I\subseteq V$. Thus, $V$ contains all but finitely many of the intervals in $\mathscr{I}_C$. Moreover, if $I=(a,b)\in\mathscr{I}_C$ is not a subset of $V$, then $V\cap I$ is open in $I$, and there are $c,d\in I$ such that $(a,c)\cup(d,b)\subseteq V$. Conversely, it’s not hard to see that any subset $V$ of $[0,1]$ that contains $C$ and all but finitely many members of $I$ and has the property described in the previous sentence, then $V$ is an open nbhd of $C$, and $\pi_C[V]$ is an open nbhd of $p_C$.
$X_C$ is therefore homeomorphic to the following space.
I’ll leave it to you to show that $X_A$ is also homeomorphic to $X$; you can pretty much just substitute $A$ for $C$ throughout the argument for $X_C$.