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.
 A: 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 $(1/n)_{n>0}$ converges to $0$ in $\mathbb{R}$ with the usual metric, so $(f(1/n))_{n>0}$ should converge to $f(0)$ in $\mathbb{R}_d$. But in a discrete space only the eventually constant sequences converge.
A: HINT: Consider any point $x$; $\{x\}$ is an open set in the discrete metric, use the openness criteria.
If you need more: 

Let $\mathbb R$ be the reals with the usual metric, $\mathbb R^\dagger$ with the discrete metric. If $\phi: \mathbb R \to \mathbb R^\dagger$ is a homeomorphism, then the preimage of $\{x\}$ is a single point (injectivity), and that single point must be an open set in the usual metric.

A: Every set in the discrete metric is open. But there exist non open sets in the usual metric.
A: Suposse that $(\mathbb R,d_1)$ and $(\mathbb R,d_2)$  are homeomorphic ($d_1$ usual metric and $d_2$ discret metric), then there exists a homeomorphism $f:(\mathbb R,d_1)\rightarrow(\mathbb R,d_2)$. Let $y\in\mathbb R$, consider the open set $\{y\}$ in $(\mathbb R,d_2)$, as $f$ is bijective, we have that $f^{-1}(\{y\})=\{x\}$ is a open set in $(\mathbb R,d_1)$. Thus we have that $(\mathbb R,d_1)$ is a discret metric space, which contradicts that $d_1$ is the usual metric.
A: 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.
