Proving $T_0$ is a topological property

Question

Let $(X,\tau)$ and $(Y,\tau_1)$ be homeomorphic topological spaces. Show if $(X,\tau)$ is a $T_0$ space then $(Y,\tau_1)$ is $T_0$ space.

My attempt:

If $(X,\tau)$ is a $T_0$ space and $x,y\in X$ $\exists\mathscr{U}$ and $\mathscr{V}$ such that $x\in\mathscr{U}$ and $y\in\mathscr{V}$ such that $\mathscr{U}\cap\mathscr{V}=\emptyset$. It could be also inteperted as $|\mathscr{U}\cap\{x,y\}|=1$.

If $f:X\to Y$ defines a homeomorphism, then $f(\mathscr{U})\in\tau_1$ once $\mathscr{U}\in\tau$. Since $x\in\mathscr{U}$ and $f$ is a bijection then $f(x)\in f(\mathscr{U})$. By the same argument $f(y)\not\in f(\mathscr{U})$. Then $|f(\mathscr{U})\cap\{f(x),f(y)\}|=1$.

Question:

Is my proof right? If not. Which are the mistakes?

Thanks in advance!

Answer 1

You have to start with $y_1, y_2 \in Y$ distinct. Then there are unique $x_1,x_2 \in X$ such that $f(x_1) =y_1$, $f(x_2) = y_2$ as $f$ is a bijection.

$X$ is $T_0$ so there is some open $U$ in $X$ such that, say, $x_1 \in U$, $x_2 \notin U$. But then $V:= f[U]$ is also open in $Y$ ($f$ is an open map) and so $y_1 = f(x_1) \in V$ and $f(x_2) \notin V$ (or else $x_2 \in f^{-1}[V] =U$ quod non).

So $Y$ is $T_0$.