Let $Y$ be a Hausdorff space, and $U,V \subset Y$ are homeomorphic under subspace topology. Does this imply if $U$ is open(or closed) then $V$ is open(or closed) under original topology?
I can't think of a counter-example or a straight-forward proof. However, it is obvious that if the space is not Hausdorff then it is fails.
Consider $Y=\{1,2,3,4,5,6\}$, $\tau=\{Y,\phi,(1,2,3)\}$, then subset $\{1,2,3\}$ is homemorphic to subset $\{4,5,6\}$ but $\{1,2,3\}$ is open while $\{4,5,6\}$ is not.
No, it does not. Consider the space $X=\big((0,1]\times\{0\}\big)\cup\big((0,1)\times(1,3)\big)$ with the topology that it inherits from the plane, let $U=(0,1)\times\{0\}$, and let $V=(0,1)\times\{2\}$. $U$ is open in $X$ but not closed, and $V$ is closed but not open, yet $U$ and $V$ are both homeomorphic to $\Bbb R$ with the usual topology.