It is true that the category of locally compact Hausdorff spaces is equivalent to the category of commutative $C^*$-algebras . . . with appropriately chosen morphisms.
Let $A$ and $B$ be commutative $C^*$-algebras. Then, a morphism from $A$ to $B$ is defined to be a nondegenerate homomorphism of $^*$-algebras from $\phi :A\rightarrow M(B)$, where $M(B)$ is the multiplier algebra of $B$. Here, nondegeneracy means that the the span of $\left\{ \pi (a)b:a\in A,b\in M(B)\right\}$ is dense in $M(B)$. Note that you need a bit of machinery to even make this into a category because it is not obvious a priori that composition makes sense. Nevertheless, it does work out. Proposition 1 on pg. 11 and Theorem 2 on pg. 12 of Superstrings, Geometry, Topology, and $C^*$-algebras (in fact this chapter is on the arXiv) respectively show that this forms a category and that the dual of this category is equivalent to the category of locally compact Hausdorff spaces.
If $(X,\tau)$ and $(X,\sigma)$ are such that $\sigma \subseteq \tau$ ($\sigma$ is weaker than $\tau$) then $\textrm{id}: (X, \tau) \to (X,\sigma)$ is continuous because a function is continuous iff every inverse image of an open set is open:
$$\forall O \in \sigma: \textrm{id}^{-1}[O]=O \in \tau \text{ as } \sigma \subseteq \tau$$
The function $\textrm{id}$ as a function from $X$ to $X$ is always a bijection (regardless of topologies on either side); this is just a set theory fact and has nothing to do with weaker/stronger topologies.
So if $(X,\tau)$ is compact and $\sigma \subseteq \tau$ is Hausdorff, then $\textrm{id}$ is continuous as I already explained and a bijection and so your very first theorem concludes from compactness of $\tau$ and Hausdorffness of $\sigma$ that $\textrm{id}: (X,\tau) \to (X, \sigma)$ is a homeomorphism. In particular it is an open map, so
$$\forall O \in \tau: \textrm{id}[O]=O \in \sigma \text{ or otherwise put: } \tau \subseteq \sigma$$
So in conclusion:
Let $(X,\tau)$ be compact Hausdorff and suppose $\sigma$ is a weaker topology than $\tau$ on $X$. Then if $\sigma$ is properly weaker, then $(X,\sigma)$ is not Hausdorff.
Similarly:
Let $(X,\tau)$ be compact Hausdorff and suppose $\sigma$ is a stronger topology than $\tau$ on $X$. Then if $\sigma$ is properly stronger, then $(X,\sigma)$ is not compact.
Proof of the latter is similar: $\textrm{id}: (X,\sigma) \to (X, \tau)$ is continuous as $\tau \subseteq \sigma$; we assumed the domain is compact and we know the codomain is Hausdorff so again we conclude this $\textrm{id}$ is also open and so $\sigma \subseteq \tau$ hence $\sigma=\tau$ contradicting *properly** stronger. So the assumption that $(X,\sigma)$ is compact must be false.
Best Answer
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.