It's important to realise that the phrase "non-commutative space" is not well-defined, in the sense that this name will mean different things to different people.
However, I will try to tell something more about $C^*$-algebras.
For convenience, I will stick with the unital case in this answer, but everything I say has non-unital counterparts.
Gelfand duality gives us a duality
$$\mathrm{Compact \ Hausdorff \ spaces} \leftrightarrow \mathrm{Commutative \ unital \ C^* algebras}.$$
This allows us to think of commutative unital $C^*$-algebras as being classical compact Hausdorff spaces. Extending this to non-commutative $C^*$-algebras, a $C^*$-algebra is sometimes called a "non-commutative topological space" and the theory of $C^*$-algebras is referred to as non-commutative topology. Some authors sometimes take this a bit further, and denote an arbitrary $C^*$-algebra $A$ by $A = C(\mathbb{X})$ where $\mathbb{X}$ can be thought of as an "imaginary topological space". So, as a purely formal object, $\mathbb{X}$ does not exist or does not make sense, yet we think of the $C^*$-algebra $A=C(\mathbb{X})$ as being functions on the imaginary space $\mathbb{X}$.
There is a similar duality between commutative von Neumann algebras and certain measure spaces, which is why the theory of von Neumann algebras is sometimes referred to as being 'non-commutative measure theory'.
Next, I will tell something more about "non-commutative compact topological groups" where the non-commutative is in the sense of non-commutative geometry. We have a forgetful functor
$$\mathrm{Compact \ Hausdorff \ groups}\to \mathrm{Compact \ Hausdorff \ spaces}$$
so it makes sense to ask how we can complete the correspondence
$$\mathrm{Compact \ Hausdorff \ topological \ groups} \leftrightarrow \quad ???$$
in a way that the right hand side corresponds to certain commutative unital $C^*$-algebras (with extra structure encoding the multiplication of the group).
At the end of the eighties, Woronowicz realised that the right hand side should correspond to what is now called a (unital) Woronowicz $C^*$-algebra, i.e. a unital $C^*$-algebra $A$ together with a unital $*$-homomorphism $\Delta: A \to A \otimes_{\operatorname{min}} A$ that is coassociative,
$$(\Delta \otimes \iota)\circ \Delta = (\iota \otimes \Delta)\circ \Delta,$$ and such that
$$\overline{\Delta(A)(1 \otimes A)}^{\|\cdot\|}= A \otimes_{\operatorname{min}} A = \overline{\Delta(A)(A \otimes 1)}^{\|\cdot\|}\quad (*)$$
Where does this come from? Well, given a compact Hausdorff group $X$, the multiplication $X \times X \to X$ dualises to a comultiplication
$$\Delta_X: C(X) \to C(X \times X) \cong C(X) \otimes C(X)$$
defined by $\Delta_X(f)(x,y) = f(xy)$ where $f \in C(X)$ and $x,y \in X$. The coassociativity of $\Delta_X$ corresponds to the associativity of the multiplication in $X$ and the condition $(*)$ corresponds to the fact that a group has inverses (which is why $(*)$ is referred to as 'quantum cancellation rules').
If the Woronowicz $C^*$-algebra $(A, \Delta)$ is commutative, it is necessarily of the form $(A, \Delta) \cong (C(X), \Delta_X)$ for some compact Hausdorff group $X$ and because of this, an arbitrary Woronowicz $C^*$-algebra $A$ is often denoted by $A= C(\mathbb{X})$ where $\mathbb{X}$ is again some imaginary object, which we refer to as being a compact quantum group. Again, as a formal object this does not exist or make sense, but it conveys the right intuition we have from the classical related theories to keep talking about objects $\mathbb{X}$ instead of talking about their related 'function algebras'.
I hope this answer helped a bit. Talking about imaginary objects can be somewhat counterintuitive and confusing at first.
Many other examples of this phenomenon occur in the world of (non-commutative) algebraic geometry, but I'll leave it to someone who actually knows about this to write an answer about this.
We know that $C(X)$, $C(Y)$ are unital $C^*$-algebras, which doesn't have anything to do with $X$ and $Y$ being homeomorphic. So I won't elaborate on that part.
Following the proof in your link: Suppose that $h:Y\to X$ is a homeomorphism. Define $H:C(X)\to C(Y)$ by $H(f)=f\circ h$. This is a continuous, scalar valued function on $Y$. It is continuous because it is a composition of continuous functions.
Fix $y\in Y$, $f,g\in C(X)$, and scalars $a,b$. Then $$H(af+bg)(y)=(af+bg)(y)=af(y)+bg(y)=aHf(y)+bHg(y),$$ so $H$ is linear.
We also have that for any $f,g\in C(X)$, $$[H(f)H(g)](y)=f(h(y))g(h(y))=(fg)(h(y))=H(fg)(y),$$ so $H(fg)=H(f)H(g)$.
Moreover, $$H(\overline{f})(y)=\overline{f}(h(y))=\overline{f(h(y))}=\overline{Hf(y)},$$ so $H(\overline{f})=\overline{Hf}$. This shows that $H$ is a $*$-homomorphism.
We note that $H$ is an isomorphism because its inverse $H^{-1}:C(Y)\to C(X)$ is given by $H^{-1}g=g\circ h^{-1}$. To see that this is the inverse of $H$, we note that $HH^{-1}g=g\circ h^{-1}\circ h=g$ and $H^{-1}Hf=f\circ h\circ h^{-1}=f$.
This is also an isometry. Because $h$ is a surjection, $f$ and $f\circ h$ have the same range.
Best Answer
$Y$ can be described as the Stone space of the measure algebra of $(X,\mu)$. That is, let $\Sigma$ be the $\sigma$-algebra on which $\mu$ is defined, let $N\subseteq\Sigma$ be the ideal of null sets, and let $B=\Sigma/N$ be the quotient Boolean algebra. Then $Y$ is naturally homeomorphic to the set $S$ of Boolean homomorphisms $B\to\{0,1\}$, topologized as a subspace of $\{0,1\}^B$.
To prove this, let us first recall that $Y$ can be described as the set of $*$-homomorphisms $L^\infty(X,\mu)\to\mathbb{C}$, with the topology of pointwise convergence. For each $b\in B$, there is a function $1_b\in L^\infty(X,\mu)$, and a $*$-homomorphism $\alpha$ must send $1_b$ to either $0$ or $1$ since $1_b^2=1_b$. It is then easy to see that $b\mapsto \alpha(1_b)$ is a Boolean homomorphism $B\to\{0,1\}$ (the Boolean operations on sets can be expressed in terms of ring operations on their characteristic functions). This defines a map $F:Y\to S$.
Note moreover that since simple functions are dense in $L^\infty(X,\mu)$, an element of $Y$ is determined by its values on characteristic functions $1_b$. Thus $F$ is injective. Also, $F$ is continuous, since the topology on $S$ is the topology of pointwise continuity with respect to evaluation at just the elements $1_b$. Since $Y$ and $S$ are both compact Hausdorff, it follows that $F$ is an embedding.
It remains to be shown that $F$ is surjective. Fix a homomorphism $h:B\to\{0,1\}$, and let $U=h^{-1}(\{1\})$. The idea is that we can then define a $*$-homomorphism $L^\infty(X,\mu)\to\mathbb{C}$ which maps a function $f$ to the "limit" of the values of $f$ along the ultrafilter $U$. To make this precise, given $f\in L^\infty(X,\mu)$ and $b\in B$, let $f[b]\subset\mathbb{C}$ denote the essential range of $f$ on $b$, and let $C_f=\{f[b]:b\in U\}$. Note that each element of $C_f$ is compact and nonempty. Also, $f[b\cap c]\subseteq f[b]\cap f[c]$, so $C_f$ has the finite intersection property. Thus $\bigcap C_f$ is nonempty. If $x\in \bigcap C_f$, then for any neighborhood $V$ of $x$ and any $b\in U$, $f^{-1}(V)\cap b$ is non-null. Since $U$ is an ultrafilter on $B$, this means $f^{-1}(V)\in U$. Now if we had two different points $x,y\in C_f$, they would have disjoint neighbooods $V$ and $W$, and then $f^{-1}(V)$ and $f^{-1}(W)$ would be disjoint elements of $U$. This is impossible.
Thus, we have shown that $C_f$ has exactly one point for each $f\in L^\infty(X,\mu)$. Define $\alpha(f)$ to be the unique element of $C_f$, which can also be described as the unique point $x$ such that given any neighborhood $V$ of $x$, for all sufficiently small $b\in U$, $f|_b$ takes values in $V$ almost everywhere. This description makes it easy to verify that $\alpha$ is a $*$-homomorphism, and that $\alpha(1_b)=h(b)$ for each $b\in B$. Thus $\alpha\in Y$ and $h=F(\alpha)$, so $h$ is in the image of $F$, as desired.
(Alternatively, to show $F$ is surjective, by Stone duality it suffices to show that the image of $F$ separates elements of $B$, since closed subspaces of the Stone space $S$ correspond to quotients of the algebra $B$. But by Gelfand duality, elements of $Y$ separate elements of $L^\infty(X,\mu)$, and so we're done since distinct elements of $B$ have distinct characteristic functions in $L^\infty(X,\mu)$.)