Topology: Homeomorphism between finite complement topology in $\mathbb{R}$ and one of its subspaces My class notes say that because $U=\mathbb{R}\backslash\\{x_1,x_2,..,x_n\\}$ has the same cardinality than $\mathbb{R}$, there exists a homeomorphism between: $(U,T_{cof})$ and $(\mathbb{R},T_{cof})$, where $T_{cof}$ is the finite complement topology. I initially thought that having the same cardinality is a necessary condition but is not sufficient to have an homeomorphism. Also, I can't manage to find a homeomorphism between these two.

You are right that in general having the same cardinality is not enough (e.g. $[0,1]$ vs $(0,1)$ with Euclidean topology). However in the case of finite complement topology it is enough.

For that consider any bijection $f:X\to Y$ with finite complement topology on both sides. It is continuous because the preimage of a finite set is finite. It is closed because the image of a finite set is finite. These two properties are enough to ensure that $f$ is a homeomorphism.