I have a question about Hausdorff topological space.
Question:
Let $X,Y$ be topological spaces.
If $X$ is a Polish space (i.e. $X$ is a separable and completely metrizable space.) and $Y=f(X)$ where $f:X \to Y$ is continuous and injective, then $Y$ is a Hausdorff space?
My attempt
Take $x,y \in Y$ with $x \neq y$ then there exist $a,b \in X$ such that $x=f(a),y=f(b)$. Since $f:X \to Y$ is injective, $a \neq b$. Since $X$ is Hausdorff space, there exist open sets $U,V \subset X$ such that $U \cap V = \emptyset$ and $a \in U,b \in V$. Therefore $x \in f(U),y \in f(V)$ and $\emptyset=f(U \cap V)=f(U) \cap f(V)$. But I don't know $f(U),f(V)$ are open subset of $Y$.
Do you know a counter example?
Thank you in advance.