This question is from Wayne Patty's Topology Page 175.
Prove that completely regular is a topological property.
I chose C is closed in X and p$\in X-C$, there exists f such f(p)=1 and f(c)=0 for all c$\in C$.
Let X is homeomorphic to Y so there exists a continuous function a : X->Y, such that $a^{-1}$ exists and is continuous.
and let there exists C' closed in Y and p' $\in Y-C'$ . But I am unable to find any function say m such that m(C)=0 and m(p')=1. But I am not able to find such a function and need help.
If $\alpha:Y\rightarrow X$ is an homeomorphism, then it is closed, that is, $\alpha(C)\subseteq X$ is closed for every closed $C\subseteq Y$. Also, since $\alpha$ is 1-1, if $p\in Y-C$, then $\alpha(p)\in X-\alpha(C)$, so using complete regularity of $X$, there is a continuous $i:X\rightarrow [0,1]$ such that $$i(\alpha(p))=1,\qquad i(x)=0\forall x\in\alpha(C)$$so consider $i'=i\circ\alpha:Y\rightarrow[0,1]$, this is a continuous map separating $C$ from $p$