The result that I want to prove is the following: Let's associate to each $x\in \mathbb{R}$ a number $r_{x}>0$. Then there is $M\in \mathbb{N}$ and a sequence $ \left( x_{n} \right)_{n\in \mathbb{N}}\subseteq \mathbb{R} $ of distincts points of $\mathbb{R}$ such that
$$r_{x_{n}}>\frac{1}{M}, \ \ \forall n\in \mathbb{N}.$$
It is clear that the result is true if the function $r$ is continuous, but I can't find a way to prove it in a general way. I know that I have to use the fact that $\mathbb{R}$ is uncountable. Can someone help me with a hint on how to attack this problem?
For each $M \in \mathbb N$ define $$R_M = \bigl\{x \in \mathbb R \mid r_x > \frac{1}{M} \bigr\} $$ Your statement is equivalent to saying that there exists $M \in \mathbb Z$ such that $R_M$ is infinite.
If not then $R_M$ is finite for all $M$. Using the sequence of inclusions $$R_1 \subset R_2 \subset R_3 \subset \cdots \subset R_m \subset \cdots $$ it follows, using finiteness of each $R_m$, that the set $\bigcup_{m \in \mathbb Z} R_m$ is countably infinite. But also $\bigcup_{m \in \mathbb Z} R_m = \{x \in \mathbb R \mid r_x > 0\} = \mathbb R$ which is uncountable.