Suppose $f:X\rightarrow Y$ is a homeomorphism. Show that if $X$ is hausdorff then so is $Y$.
My attempt: Let $y_1,y_2\in Y$ be distinct, by bijectivity of $f$, there exists distinct $x_1,x_2 \in X$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$. Since $X$ is hausdorff, there exists disjoint open subsets of $X$, $V_1$ and $V_2$ containing $x_1$ and $x_2$ respectively. Since $f$ is a homeomorphism, it is an open map, hence $f(V_1)\cap f(V_2)$ is the union of two open sets. Since $f$ is injective, $f(V_1 \cap V_2)= f(V_1) \cap f(V_2)= \varnothing$, and $f(V_1),f(V_2)$ contain $y_1,y_2$ respectively. Thus $Y$ is hausdorff.
Is it correct?