I want to show regularity is a homeomorphism invariant.
Here is my proof:
Let $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ be two topological spaces with $(X, \mathcal{T})$ being regular. Let $f: X \to Y$ be a homeomorphism, we want to show that $(Y, \mathcal{U})$ is regular. Let $y \in Y$ be a point and $D \subseteq Y$ be any closed set in $Y$ not containing $y$. Consider the point $x = f^{-1}(y) \in X$ and the set $C = f^{-1}(D)$ in $X$. Since $f$ is a homeomorphism (continuous with continuous inverse), we have that $C$ is a closed set in $X$. We prove that $C$ does not contain $x$. To see this assume the contrapositive, i.e., $x \in C$. Since $f^{-1}$ is the inverse of a continuous bijection we have that $f(x) = y$ and that $y \in f(C)$. But $f(C) = D$ and by hypothesis, we know that $y \notin D \longrightarrow$ Contradiction!. So $x \notin C$. Now, by regularity of $X$, there exist open sets $U$ and $V$ in $X$ such that $x \in U$ and $C \subseteq V$ and $U \cap V = \emptyset$. Now let's define $U' = f(U)$ and $V' = f(V)$. Now $U' \cap V' = f(U) \cap f(V) = f(U \cap V) = \emptyset$. The second last equality follows because $f$ is a bijection. Since $f^{-1}$ is continuous, $U'$ and $V'$ are also open in $Y$. Now since, $x\in U$, we have that $f(x) \in f(U)$ and $C \subseteq V$ so $f(C) \subseteq f(V)$. In other words, we have $f(x) = y \in U'$ and $f(C) = D \subseteq V'$ and $U' \cap V' = \emptyset$. Thus we have found $U'$ and $V'$ open in $Y$ such that $x \in U'$ and $D \subseteq V'$ and $U' \cap V' = \emptyset$. This shows that $(Y, \mathcal{T})$ is regular.
Is my proof correct?