How to prove that $\mathbb R$ with usual metric and $\mathbb R$ with the discrete metric are not homeomorphic. I know that two metric spaces are homeomorphic if there is a function from one to another such that $f$ is continuous, one-to-one, onto, with continuous inverse $f^{-1}$.
I know how to prove two metric spaces are homeomorphic.

I don't understand how to prove that two metric spaces are not homeomorphic.
I have to prove using knowledge I have of open sets, closed sets, limit and continuity.
I cannot use compactness, disconnectedness, etc.
Let $f\colon \mathbb{R}\to\mathbb{R}_d$ be a homeomorphism, where the domain has the usual metric and the codomain the discrete metric.

The sequence $(1/n)_{n&gt;0}$ converges to $0$ in $\mathbb{R}$ with the usual metric, so $(f(1/n))_{n&gt;0}$ should converge to $f(0)$ in $\mathbb{R}_d$. But in a discrete space only the eventually constant sequences converge.
Consider connectedness. Alternatively, one space is separable, and the other is not. Alternatively, the convergent sequences in one space are eventually constant, and you can count how many there are.