Prove that if $h$ is a homeomorphism from $(X,\mathscr{S})$ to $(Y,\mathscr{T})$, then $\forall a, b \in X \left( a \trianglelefteq_{\mathscr{S}} b \iff h(a) \trianglelefteq_{\mathscr{T}} h(b) \right)$.
Furthermore, the preorder relation $\trianglelefteq_{\mathscr{T}}$ is defined as such: $\trianglelefteq_{\mathscr{T}} = \{(a,b) \in X^2 : a \in \overline{\{b\}}\}$.
At first, I tried to use the definition of homeomorphism as follows:
Let $h$ be a homeomorphism from $(X,\mathscr{S})$ to $(Y,\mathscr{T})$. Now by hypothesis, $a \trianglelefteq_{\mathscr{S}} b $ means that $a \in \overline{\{b\}}$ within the topology $\mathscr{S}$. Since $h$ is a homeomorphism from $(X,\mathscr{S})$ to $(Y,\mathscr{T})$, then we have a one-to-one correspondence from $(X,\mathscr{S})$ onto $(Y,\mathscr{T})$. We cannot choose any particular bijection (at least I don't think so since we are trying to show this holds for all $a,b \in X$), then we start from $a \trianglelefteq_{\mathscr{S}} b$ since this implies that $a \in \overline{\{b\}}$. Applying $h$, we get $h(a) \in h(\overline{\{b\}})$ and this completes this direction.
For the other direction, we are assuming that $h(a) \trianglelefteq_{\mathscr{T}} h(b)$ which means that $h(a) \in h(\overline{\{b\}})$. Since $h$ is a homeomorphism, then $h^{-1}$ is also continuous. Applying this to our hypothesis gives us that $a \trianglelefteq_{\mathscr{S}} b$ as desired.
I do not feel too convinced about this proof. Perhaps I am misunderstanding what needs to be shown or maybe I am not using the definitions correctly. Can somebody please point me in the right direction in order to write a correct proof? Many thanks in advance, your help is appreciated.